丁香园AV

We first note that the domain of \(P\) is \(q \in [0,\infty)\text{.}\) We differentiate twice:

To find any critical points of \(P\text{,}\) we first set

Next, we look for any points where \(P'(q)\) is undefined, which gives no solutions. Hence, the only critical point of \(P\) is \(q=130\text{.}\) Since \(P''(130)\lt 0\text{,}\) by the Second Derivative Test for relative extrema, \(P\) has a relative maximum at \((130, P(130)) = (130, 69500)\text{.}\)

Furthermore, since \(P\) is an upside down parabola, this relative maximum must be the absolute maximum. Hence, the maximum monthly profit is $69,500 per month. This maximum occurs when 130 units are produced.

The domain of the profit function must be \([0,\infty)\text{.}\) We now differentiate:

To find the critical points of \(P\text{,}\) we set \(P'(q)=0\text{:}\)

Since \(q=0\) is an endpoint, it cannot be a critical point. Hence, the profit function has one critical point at \(q=20/3\text{.}\) We further notice that there are no points where \(P'(q)\) DNE, and so this gives no additional critical points.

We now compare profits:

We further notice that the profit is decreasing for all \(q > 20/3\text{.}\) We conclude that the absolute maximum profit the manufacturer realizes is $29.63 per day when the level of production is set to approximately 7 units.

We first construct the daily profit as a function of \(q\text{.}\)

To determine the domain of \(P\text{,}\) we first notice that we require both the quantity demanded \(q\) and the unit price \(p\) to be positive:

Therefore, the domain is \([0, 25000]\text{.}\)

Next, we find the interior critical points of \(P\) by computing

which gives a single solution, \(q=6000\) (note that this point is in the domain of \(P\)). Since there are no points \(q\) in the domain of \(P\) for which \(P'(q)\) does not exist, this must be the only critical point.

We have defined the profit function on a closed interval, and so to determine the absolute maximum, we can compare the value of the profit function at all critical points and endpoints:

So the absolute maximum is attained when \(q=6000\text{.}\)

Therefore, with a production level of \(6000\) units per day, the profit is maximized at $17,600.

We first construct the weekly profit function, \(P(q)\text{:}\)

where \(q \geq 0\text{.}\) Now calculate

Since \(P'(q)\) DNE has no solutions, we set \(P'(q) = 0\) to find any critical points.

Using the quadratic formula, we find two solutions, \(q = 215.25\text{,}\) and \(q=-1548.58\text{.}\) We reject the second root (since it is not in the domain of \(P\)).

We now compare the profit at the critical point and at the endpoints:

Therefore, \(q \approx 215\) is the absolute maximum of \(P\text{.}\)

Therefore, the profit is maximal when the level of production is at approximately \(215\) units per week.

We first determine the domain of the demand equation:

We also know that \(q\) must be a positive number. Therefore, the revenue function is given by

We now look for critical points of the revenue function. We compute

The only point in the domain of \(R\) for which \(R'\) is undefined is when \(q=600\text{.}\) Since this is a boundary point, it cannot be a critical point. We therefore need ot solve \(R'(q) = 0\text{:}\)

To determine the absolute maximum revenue, we compare

We conclude that in order to maxmize revenue, \(400\) units should be manufactured and sold.

We define

for \(q > 0\text{.}\) We first differentiate the average revenue:

Notice that \(\overline{R}'(q)\) is undefined when \(q=0\text{,}\) but since this point is not in the domain of \(\overline{R}(q)\text{,}\) we reject this as a critical point. Therefore, the only critical point will occur when

Notice that this last equality can be written as

Hence, the critical point of the average revenue function satisfies

Differentiating once more, we see that

Therefore, when \(qR'(q) = q\text{,}\) we see that

Since \(R(q)\) is concave down everywhere and \(q>0\text{,}\) this means that the average revenue function is concave down when \(\overline{R(q)} = R'(q)\text{.}\) By the Second Derivative Test, this critical point is a relative maximum.

We compute

We now solve for the critical points of \(G\text{:}\)

Since \(t=0\) is an endpoint, it is not a critical point. Furthermore, since \(G'(t)\) is defined for all \(t\) in the domain of \(G\text{,}\) we conclude that there \(G\) has a single critical point at \(t=8\text{.}\)

We now compare the value of \(G\) at the critical point and both boundary points:

Therefore, the GDP is at its highest after 8 years.

We differentiate:

Setting \(a'(t)=0\) gives \(t=42.1814\) as the only real solution. This is in the domain of \(a(t)\) and is therefore a critical point of \(a\text{.}\) Setting \(a'(t)\) DNE gives no solutions.

We now compare:

We conclude that the value of the account is maximal between the 42nd and 43rd years.

We differentiate with respect to \(t\text{:}\)

To find the critical points, we set \(N'(t)=0\text{.}\) This gives three solutions: \(t=0.812897\text{,}\) \(t=3.0978\text{,}\) \(t=5.00359\text{.}\) In considering the domain of \(N\text{,}\) we reject the last solution. So we compare:

We conclude that over the 5 year period, the number of bakeries was maximal after about \(0.8\) years, and minimal after abut \(3\) years.

We first construct the average cost function.

for \(q > 0\text{.}\) Differentiating, we find that

Since there are no points in the domain of \(C\) for which \(\overline{C}'(q)\) is undefined, it remains to set \(\overline{C}'(q)=0\text{:}\)

We consider the following sign diagram for \(\overline{C}'(q)\text{:}\)

Since the average cost is decreasing on the interval \((0,86.9784)\text{,}\) and increasing from \((86.9784,\infty)\text{,}\) we conclude that the absolute minimal average cost occurs when about 20,000 units are produced.

Suppose we have the following rectangle:

Then the area is given by

and the perimeter is

where we have that \(x,y \geq 0\text{.}\)

We will use the constraint on the perimeter to write \(y\) as a function \(x\) :

Therefore, we can write the area as a function of only one variable:

with domain \(0\leq x\leq 50\text{.}\) Differentiating, we find

This means that \(x=25\) is the only critical point. Now compare:

We conclude that the area is maximized when the dimensions are \(25\)x\(25\text{.}\)

Suppose we have the following rectangle:

Then the area is given by

and the perimeter (for some fixed constant \(P\)) is

where we have that \(x,y \geq 0\text{.}\)

We will use the constraint on the perimeter to write \(y\) as a function \(x\) :

Therefore, we can write the area as a function of only one variable:

with domain \(0\leq x\leq \frac{P}{2}\text{.}\) Differentiating, we find

This means that \(x=P/4\) is the only critical point. Now compare:

We conclude that the area is maximized when the dimensions are \(\dfrac{P}{4}\)x\(\dfrac{P}{4}\text{.}\)

Suppose the base of the box has side length\(l\) and the height of the box is \(h\text{.}\) Then the volume of the box is

and the surface area is

If the volume of the box is constrained, then we can write

And so we can write

where \(l > 0\text{.}\)

Differentiating, we find

Therefore, the critical point of the surface area is \(l=2\cdot 5^{2/3}\text{.}\)

We construct the following sign diagram for \(S'\text{:}\)

Therefore, the surface area is decreasing for \(l\in(0,2\cdot 5^{2/3})\) and increasing for \(l \in (2\cdot 5^{2/3},\infty)\text{.}\) We conclude that the surface area is minimised with dimensions \(2\cdot 5^{2/3}\) x \(2\cdot 5^{2/3}\) x \(5^{2/3}\text{.}\)

Suppose the dimensions of the base of the box are \(l\)x\(l\text{,}\) and the height of the box is \(h\text{.}\) The volume is then

and the surface area (noting that the top and bottom are doubled) is

Since the volume is constrained at \(V=200\text{,}\) we can write

which means that

where \(l > 0\text{.}\)

Differentiating \(S\) with respect to \(l\text{,}\) we compute

Therefore,

We construct the following sign diagram for \(S'\text{:}\)

We conclude that the surface area is minimized with dimensions \(10^{2/3}\) x \(10^{2/3}\) x \(2\cdot 10^{2/3}\text{.}\)

Suppose the base of the box has side length\(l\) and the height of the box is \(h\text{.}\) Then the volume of the box is

and the surface area is

If the volume of the box is constrained, then we can write

And so we can write

where \(l > 0\text{.}\)

Differentiating, we find

Therefore, the critical point of the surface area is when \(l = (2V)^{1/3}\text{.}\)

We construct the following sign diagram for \(S'\text{:}\)

Therefore, the surface area is minimised with dimensions \((2V)^{1/3}\) x \((2V)^{1/3}\) x \(h\text{,}\) where

This gives a ratio

We draw the following diagram, where the blue lines indicate the fencing (\(x\) and \(y\) are measured in feet):

The area will then be

with perimeter

Since there is a fixed amount of fencing material, we require

Hence, we can write the area as a function of \(x\) only:

for \(0 \leq x \leq 50\text{.}\) Now differentiate:

and so \(x = 25\) gives the only critical point. Now compare:

We conclude that 1250 ft\(^2\) is the maximal area.

We draw the following diagram, where the blue lines indicate the fencing (\(x\) and \(y\) are measured in feet):

The area will then be

with perimeter

Since there is a fixed amount of fencing material, we require

Hence, we can write the area as a function of \(x\) only:

for \(0 \leq x \leq l/2\text{.}\) Now differentiate:

and so \(x = l/4\) gives the only critical point. Now compare:

We conclude that \(\dfrac{l^2}{8}\) ft\(^2\) is the maximal area.

Let \(q\) be the number of units in question. If our fixed costs total $3000 and our marginal cost is $2 per unit, then our total cost function \(C(q)\) (in dollars) is given by

Now, let \(p\) be the unit price (in dollars). Then marketing tells us that we can sell \(q=500\) units for each dollar below $10 that we set the price. Let \(0 \le h \le 10\text{.}\) Then

for some \(h \in [0,10]\text{.}\) We need to construct our revenue \(R\) as a function of \(q\text{,}\) and so we let \(h=q/500\) (where we rearranged the second equation above). Then,

Finally, we construct our profit function \(P\) as a function of \(q\) units sold.

where \(0\leq q \leq 5000\text{.}\)

Now, find the critical points of \(P\text{.}\)

which gives \(q=2000\) as our only interior critical point. Comparing

we conclude that the profit is at an absolute maximum when we sell \(2000\) units. Since \(q=2000\text{,}\) this means that \(h = 4\) and hence \(p = 10-4 = 6.\) Therefore, each unit is sold at $6 per unit. The profit yielded will be $5000.

We first draw a sketch of the scenario. The shaded area is the area to be optimized.

Therefore,

We wish to write area as a function of \(w\) only. So let

Hence,

where \(0 \leq w \leq 200\text{.}\)

We can now differentiate with respect to \(w\text{:}\)

And so \(A'(w)=0\) when \(w= \pm \sqrt{2} 100\) and \(A'(w)\) is undefined when \(w = \pm 200\text{.}\) Since the domain of \(A\) is \(0 \leq w \leq 200\text{,}\) we conclude that there is only one interior critical point at \(w = \sqrt{2} 100\text{.}\) We now compare the value of \(A\) at this critical point with value of \(A\) at the boundary points:

Therefore, the maximum area of the rectangle is \(100\text{.}\)

We first draw a sketch of the scenario. The shaded area is the area to be optimized.

Therefore,

We wish to write area as a function of \(w\) only. So let

Hence,

where \(0 \leq w \leq 2r\text{.}\)

We can now differentiate with respect to \(w\text{:}\)

And so \(A'(w)=0\) when \(w= \pm \sqrt{2} r\) and \(A'(w)\) is undefined when \(w = \pm 2r\text{.}\) Since the domain of \(A\) is \(0 \leq w \leq 2r\text{,}\) we conclude that there is only one interior critical point at \(w = \sqrt{2} r\text{.}\) We now compare the value of \(A\) at this critical point with value of \(A\) at the boundary points:

Therefore, the maximum area of the rectangle is \(r^2\text{.}\)

Let \(r\) be the radius and \(h\) be the height. Then the volume of the cylinder is

and the total surface area (including top and bottom) is

Constraining the surface area allows us the write \(h\) as a function of \(r\text{:}\)

Hence, as a function of the radius, we write

Notice that the domain of \(V(r)\) is \(0 \lt r \leq \frac{5}{\sqrt{\pi}}\text{.}\)

We can now differentiate the volume with respect to \(r\text{:}\)

Setting \(V'(r) = 0\) gives \(r = \pm \frac{5}{\sqrt{3\pi}}\text{.}\) We reject the negative solution, and so are left with only one critical point. To determine the absolute maximum, we compare values:

We conclude that the maximum volume is achievable with a height to radius ratio of

Let \(r\) be the radius and \(h\) be the height. Then the volume of the cylinder is

and the total surface area (including top and bottom) is

Constraining the surface area allows us the write \(h\) as a function of \(r\text{:}\)

Hence, as a function of the radius, we write

Notice that the domain of \(V(r)\) is \(0 \lt r \leq \frac{\sqrt{S/2}}{\sqrt{\pi}}\text{.}\)

We can now differentiate the volume with respect to \(r\text{:}\)

Setting \(V'(r) = 0\) gives \(r = \pm \frac{\sqrt{S/2}}{\sqrt{3\pi}}\text{.}\) We reject the negative solution, and so are left with only one critical point. To determine the absolute maximum, we compare values:

We conclude that the maximum volume is achievable with a height to radius ratio of

We make the following diagrams illustrating the amount of construction material required per container. For the top and bottom, we need:

And for the side of the cylinder, we need:

Suppose \(r\) and \(h\) are measured in cm. Then the total surface area of construction material (in cm\(^2\)) is

We further have the constraint on the volume: (note that 1L = 1000 cm\(^3\))

Therefore, we write (for \(r \neq 0\))

and so we can write the surface area as a function of radius only:

Differentiating with respect to \(r\text{,}\) we find

Notice that \(S'(r)\) is undefined at \(r=0\text{,}\) but this point is not in the domain of \(S\) and so is not a critical point. Setting \(S'(r)=0\) gives \(r=5\) as the only critical point.

Notice that

and so \(S''(r) > 0\) for all \(r > 0\text{.}\)

Since \(S\) is concave up on its entire domain, and has a single critical point at \(r=5\text{,}\) this must be the absolute minimum. We conclude that the amount of construction material is minimized when \(r=5\) cm and \(h= \dfrac{40}{\pi}\) cm. This gives a ratio of

We make the following diagrams illustrating the amount of construction material required per container. For the top and bottom, we need:

And for the side of the cylinder, we need:

Suppose \(r\) and \(h\) are measured in some units. Then the total surface area of construction material (in units\(^2\)) is

We further have the constraint on the volume:

where \(V\) is some positive, nonzero constant and is measured in units\(^3\text{.}\) Therefore, we write (for \(r \neq 0\))

and so we can write the surface area as a function of radius only:

Differentiating with respect to \(r\text{,}\) we find

Notice that \(S'(r)\) is undefined at \(r=0\text{,}\) but this point is not in the domain of \(S\) and so is not a critical point. Setting \(S'(r)=0\) gives \(r=\frac{1}{2}\sqrt[3]{V}\) as the only critical point.

Notice that

and so \(S''(r) > 0\) for all \(r > 0\text{.}\)

Since \(S\) is concave up on its entire domain, and has a single critical point at \(r=\frac{1}{2}\sqrt[3]{V}\text{,}\) this must be the absolute minimum. We conclude that the amount of construction material is minimized when \(r=\frac{1}{2}\sqrt[3]{V}\) cm and \(h= \dfrac{4\sqrt[3]{V}}{\pi}\) cm. This gives a ratio of

We make a sketch of the scenario described:

Both cones are right circular cones, so if we let \(H\) and \(R\) be the height and radius of the outer (larger) cone, respectively, with \(h\text{,}\) \(r\) being the height and radius of the inner (smaller) cone, we must have:

Then, by similar triangles,

for \(0 \lt r \lt R\) and \(0 \lt h \lt H\text{.}\) Notice that if \(h=H\) we would require \(r=0\text{,}\) and similarly if \(r=R\) we would require \(h=0\text{.}\) Similarly, taking either \(r\) or \(h\) equal to zero would result in a volume of zero.

\(H\) and \(R\) are fixed and nonzero, so we can write

The volume of the inner cone is

and by using the above relation, we have for \(0 \lt h \lt H\text{,}\)

Now differentiate with respect to \(h\text{:}\)

Solving \(V'(h) = 0\) gives

The solution \(h=H\) is not in the domain of \(V\text{,}\) and so we get one critical point at \(h = \frac{H}{3}\text{.}\) The corresponding radius is \(r = \frac{2R}{3}\text{.}\) Hence, the maximum volume of the inner cone is

Since the volume of the outer cone is

we have

The container is made up of two parts:

Therefore, the total volume of the container (for some fixed, nonzero constant \(V\)) is

This means that we can write

where \(0 \lt r \lt \sqrt[3]{\frac{3V}{2\pi}}\text{.}\)

1. If the container is to be made with no bottom, then the total surface area is

   \begin{equation*} S(r,h) = 2\pi r h + 2\pi r^2\text{.} \end{equation*}

   If the cost of the top material is twice the cost of the side material, then we want to minimize the cost function:

   \begin{equation*} C(r,h) = 2\pi rh + 4\pi r^2\text{.} \end{equation*}

   Using the relation \(h=h(r)\text{,}\) we write (for \(r>0\))

   \begin{equation*} C(r) = \frac{2V}{r} + \frac{16}{3} \pi r^2\text{.} \end{equation*}

   Differentiating with respect to \(r\) gives

   \begin{equation*} C'(r) = -\frac{2V}{r^2} + \frac{32}{3}\pi r\text{.} \end{equation*}

   We set \(C'(r) = 0\) to find one critical point (remember that \(r=0\) is not in the domain of \(C\)):

   \begin{equation*} r = \sqrt[3]{\frac{3V}{16\pi}}\text{.} \end{equation*}

   At this value of \(r\text{,}\) the corresponding height is

   \begin{equation*} h = \frac{V-\frac{2}{3}\pi r^3}{\pi r^2} = \frac{7V}{8\pi \left(\frac{3V}{16\pi}\right)^{2/3}}\text{.} \end{equation*}

   This gives a ratio of

   \begin{equation*} \frac{r}{h} = \frac{3V \cdot 8\pi}{16 \pi \cdot 7V} = \frac{3}{14}\text{.} \end{equation*}

2. Now if the container is to be made with a circular bottom, the total surface area is

   \begin{equation*} S(r,y) = 2\pi r h + 2\pi r^2 + \pi r^2\text{.} \end{equation*}

   Hence, if the material cost for the top is twice the cost of the side material, and the cost for the bottom is 1.5 times to cost of the side material, we can construct the following cost function:

   \begin{equation*} C(r,h) = (1) \cdot 2\pi r h + (2) \cdot 2\pi r^2 + (1.5) \cdot \pi r^2 = 2\pi rh +\frac{11}{2} \pi r^2\text{.} \end{equation*}

   So as a function of \(r\) only, we find

   \begin{equation*} C(r) = \frac{2V}{r} + \frac{25}{6} \pi r^2\text{.} \end{equation*}

   Now differentiate:

   \begin{equation*} C'(r) = -\frac{2V}{r^2} + \frac{25}{3} \pi r\text{.} \end{equation*}

   Setting \(C'(r) = 0\) gives

   \begin{equation*} r = \sqrt[3]{\frac{6V}{25\pi}}\text{.} \end{equation*}

   At this value of \(r\text{,}\) the corresponding value of \(h\) is

   \begin{equation*} h = \frac{13V}{25\pi \left(\frac{6V}{25\pi}\right)^{2/3}}\text{.} \end{equation*}

   This gives a ratio of

   \begin{equation*} \frac{r}{h} = \frac{6}{13}\text{.} \end{equation*}

The piece of cardboard is to be cut as follows:

This results in a box with dimensions \((1-2x) \times (0.5-2x) \times x\text{.}\) The volume of this box is then

for \(0 \leq x \leq 0.25\text{.}\) We differentiate:

and so \(V'(x) = 0\) for \(x= 1/4 \pm 1/\sqrt{48}\text{.}\) Only one of these critical points is in the domain of \(V\text{:}\)

Now compare:

We conclude that the dimensions which maximize the volume of the box are

We start with the general case, where we have the following scenario: The piece of cardboard is to be cut as follows:

Assume that \(b \leq a\text{.}\) Then \(0 \leq x \leq \frac{b}{2}\text{.}\) The volume of the resulting box is therefore

Differentiating, we find

Now set \(V'(x) = 0\text{,}\) which gives

1. If we have \(a=b\text{,}\) then

   \begin{equation*} x = \frac{1}{6} \bigl(\pm \sqrt{b^2 - b^2 + b^2} + b + b\bigr)= \frac{1}{6} \bigl( 2b \pm b\bigr)\text{.} \end{equation*}

   Hence, \(x=\frac{b}{6}\) is the only critical point. We compare:

   \begin{equation*} V(0) = 0, \ \ V(b/6) = \frac{2b^3}{27}, \ \ V(b/2) = 0\text{.} \end{equation*}

   We conclude that \(x = \frac{b}{6}\) is the optimal size of the square.

2. Now let \(a \neq b\text{.}\) Since \(x \leq b/2\text{,}\) and we took \(b \lt a\) we see that

   \begin{equation*} x = \frac{1}{6} \bigl(-\sqrt{a^2 - a b + b^2} + a + b\bigr) \end{equation*}

   is the only critical point. Since

   \begin{equation*} V(0) = 0, \ \ \text{ and } \ \ V(b/2) = 0\text{,} \end{equation*}

   we conclude that this choice of \(x\) must maximize the volume.

We make the following diagram, where \(0 \leq w \leq 1.5\) m:

Therefore, the surface area of the window is the surface area of the top part:

plus the surface area of the rectangular part:

So if the top part only transmits half as much light as the rectangular part, we can construct the following function which is to be maximized:

where \(0 \leq w \leq 1.5\text{.}\) We differentiate:

So \(I'(w) = 0\) when \(w = \frac{8}{4-\pi}\text{.}\) This is not in the domain of \(I\text{,}\) and so there are no critical points. So we compare:

We conclude that the amount of light let in is maximized when the width of the window is \(1.5\) m. We make the following diagram:

Therefore, the surface area of the window is the surface area of the top part:

plus the surface area of the rectangular part:

So if the top part only transmits only \(0 \leq k \leq 1\) times as much light as the rectangular part, we can construct the following function which is to be maximized:

where \(0 \leq w \leq 2H\text{.}\) We differentiate:

We set \(I'(w)=0\) to find any critical points:

Therefore, if \(k \geq 2/\pi\text{,}\) there is no critical point. But for \(0 \leq k \lt \frac{2}{\pi}\text{,}\) there is a critical point at \(w = \dfrac{H}{1-\frac{k\pi}{2}}\text{..}\)

Let's first consider \(k \geq 2/\pi\text{:}\) then we compare

Therefore, the window which lets in the most amount of light is a semicircle (i.e. there should be no rectangular part).

Now if \(0 \leq k \lt 2/\pi\text{,}\) we find that the light intensity is maximized when \(w = \dfrac{H}{1-\frac{k\pi}{2}}\text{.}\)

We make a sketch of the scenario described:

The total area of the poster board is therefore

We want to minimize \(f\) subject to the constraint \(A=x\times y\text{.}\) We use this constraint to write \(f\) as a function of \(x\) only:

Now differentiate:

Therefore, solving \(f'(x)=0\) gives

Since \(x\) is a side-length, we reject the negative solution. Therefore, there is a single critical point at \(x=\sqrt{Ab/a}\text{.}\) Since

which is positve for all \(x>0\text{,}\) we conclude that this critical point is the absolute minimum.

Therefore, the area of the poster board is minimized with a ratio of

Consider the following cross-section of a cylinder of radius \(r_c\) and height \(h\) inside a sphere of radius \(r_s\text{:}\)

Then we see that, in order for the cylinder to fit inside the sphere, we must have

The volume of the cylinder is

and so we can use the constraint on \(r_c\) to write

This formula is only valid for \(0 \leq h \leq 2r_s\text{.}\)

Differentiating, we find

Setting \(V_c'(h) = 0\text{,}\) we find

to be the only critical point of \(V_c\) (we reject the negative solution). We now compare the value of \(V_c\) at this critical point and at the boundary points:

Therefore, the maximum possible volume of the cylinder is \(\pi r_s^2\text{.}\) The ratio of the volume of this cylinder to the volume of the sphere is therefore

Let \(l\) be the length of the box and \(w\) be the width of the box in inches. If the box has square front and back faces, then the volume of this box must be

where we require \(l + 4w \leq 108\text{.}\) In order to maximize the volume, let

Then we can rewrite the volume as a function of length only:

We now differentiate with respect to \(l\text{:}\)

which gives one interior critical point at \(l=36\text{.}\) We now compare volumes:

Hence, the volume of the box is maximal with dimensions \(36 \times 18 \times 18\) in.