丁香园AV

Classify the critical points of the following functions:

1. \(\displaystyle f(x,y)=x^2+4y^2-2x+8y-1\)

   Answer
   minimum at \((1,-1)\)
   Solution

   The first-order partial derivatives are

   \begin{equation*} f_x = 2x-2, \ \ \text{ and } \ \ f_y= 8y+8\text{.} \end{equation*}

   To find the critical points of \(f\text{,}\) we set both \(f_x=0\) and \(f_y=0\text{:}\)

   \begin{equation*} f_x = 0 \implies x = 1, \ \text{ and } f_y = 0 \implies y = -1\text{.} \end{equation*}

   This gives exactly one critical point at \((1,-1)\text{.}\) To classify this critical point, we construct the discriminant:

   \begin{equation*} D(x,y) = f_{xx}f_{yy} - \left(f_{xy}\right)^2 = 2(8)-0 = 16 > 0\text{.} \end{equation*}

   Since the discriminant is positive everywhere on its domain, and \(f_{xx} > 0\text{,}\) the point \((1,-1)\) must be a relative minimum.

2. \(\displaystyle f(x,y)=x^2-y^2+6x-10y+2\)

   Answer
   none
   Solution

   The first-order partial derivatives of \(f\) are

   \begin{equation*} f_x = 2x+6, \ \ \text{ and } f_y = -2y-10\text{.} \end{equation*}

   To find the critical points of \(f\text{,}\) we set both \(f_x=0\) and \(f_y=0\text{:}\)

   \begin{equation*} f_x = 0 \implies x = -3, \ \ \text{ and } f_y = 0 \implies y=-5\text{.} \end{equation*}

   Therefore, there is only one critical point at \((-3,-5)\text{.}\) To classify this critical point, we construct the discriminant:

   \begin{equation*} D(x,y) = f_{xx}f_{yy} - \left(f_{xy}\right)^2 = 2(-2)-0 = -4 \lt 0\text{.} \end{equation*}

   Since the discriminant is negative everywhere in its domain, the function \(f\) has no relative extrema.

3. \(\displaystyle f(x,y)=xy\)

   Answer
   none
   Solution

   The first-order partial derivatives of \(f\) are

   \begin{equation*} f_x = y, \ \ \text{ and } \ \ f_y = x\text{.} \end{equation*}

   To find the critical points of \(f\text{,}\) we set both \(f_x=0\) and \(f_y=0\text{:}\)

   \begin{equation*} f_x = 0 \implies y=0, \ \ \text{ and } f_y = 0 \implies x=0\text{.} \end{equation*}

   Thus, the point \((0,0)\) is the only critical point of \(f\text{.}\) To classify this critical point, we construct the discriminant:

   \begin{equation*} D(x,y) = f_{xx}f_{yy} - \left(f_{xy}\right)^2 = 0\text{.} \end{equation*}

   Therefore, the discriminant tells us nothing about the behaviour of \(f\) at this critical point. Instead, we notice that the function \(f(x,y) = xy\) is increasing in some directions as we move away from \((0,0)\text{,}\) but decreasing in others. Therefore this must be a saddle.

4. \(\displaystyle f(x,y)=9+4x-y-2x^2-3y^2\)

   Answer
   maximum at \((1,-1/6)\)
   Solution

   We find the first partial derivatives of \(f(x,y)\) to be

   \begin{equation*} f_x = 4 - 4x, \ \ \text{ and } \ \ f_y = -1-6y\text{.} \end{equation*}

   To find the critical points of \(f\text{,}\) we set both \(f_x=0\) and \(f_y=0\text{:}\)

   \begin{equation*} f_x = 0 \implies x=1, \ \ \text{ and } f_y = 0 \implies y=-1/6\text{.} \end{equation*}

   Therefore, the point \((1, -1/6)\) is the only critical point of \(f\text{.}\) To classify this critical point, we construct the discriminant:

   \begin{equation*} D(x,y) = f_{xx}f_{yy} - \left(f_{xy}\right)^2 = -4(-6) - 0 > 0\text{.} \end{equation*}

   Since the discriminant is positive and \(f_{xx} \lt 0\) everywhere, we conclude that the critical point \((1,-1/6)\) is a relative maximum.

5. \(\displaystyle f(x,y)=x^2+4xy+y^2-6y+1\)

   Answer
   none
   Solution

   We find the first partial derivatives of \(f(x,y)\) to be

   \begin{equation*} f_x=2x+4y, \ \ \text{ and } \ \ f_y=4x+2y-6\text{.} \end{equation*}

   To find the critical points of \(f\text{,}\) we set \(f_x=0\) and \(f_y=0\text{:}\)

   \begin{equation*} 2x+4y = 0 \implies x=-2y\text{,} \end{equation*}

   so then

   \begin{equation*} 4x+2y-6=0 \implies 4(-2y)+2y-6 = 0 \implies y=-1\text{,} \end{equation*}

   which in turn gives \(x=2\text{.}\) Therefore \((2,-1)\) is the only critical point. To classify this point, we construct the discriminant,

   \begin{equation*} D(x,y) = f_{xx}f_{yy}-\left(f_{xy}\right)^2 = (2)(2)-4^2 \lt 0 \end{equation*}

   everywhere and in particular at \((2,-1)\text{.}\) Thus, \(f\) has no relative extrema.

6. \(\displaystyle f(x,y)=x^2-xy+2y^2-5x+6y-9\)

   Answer
   minimum at \((2,-1)\)
   Solution

   The first-order partial derivatives of \(f\) are

   \begin{equation*} f_x = 2x - y - 5, \ \ \text{ and } \ \ f_y = -x+4y+6\text{.} \end{equation*}

   To find the critical points of \(f\text{,}\) we set both \(f_x=0\) and \(f_y=0\text{:}\)

   \begin{equation*} f_x = 0 \implies y = 2x-5\text{.} \end{equation*}

   Plugging this equation into \(f_y = 0\text{,}\) we find

   \begin{equation*} -x + 4(2x-5)+6 = 0 \implies x = 2\text{.} \end{equation*}

   When \(x=2\text{,}\) we must have \(y=2(2)-5) = -1\text{.}\) Hence, there is excatly one critical point at \((2,-1)\text{.}\) To classify this point, we construct the discriminant,

   \begin{equation*} D(x,y) = f_{xx}f_{yy}-\left(f_{xy}\right)^2 =2(4)-(-1) = 9 > 0\text{,} \end{equation*}

   which is positive everyone on the domain of \(f\text{.}\) Since we further have that \(f_{xx} > 0\) everywhere, \(f\) must have a relative minimum at the point \((2,-1)\text{.}\)

7. \(\displaystyle f(x,y)=xy(5x+y-15)\)

   Answer
   minimum at \((1,5)\)
   Solution

   The first-order partial derivatives of \(f\) are

   \begin{equation*} f_x = 10xy + y^2 - 15y, \ \ \text{ and } \ \ f_y = 5x^2+2xy-15x\text{.} \end{equation*}

   To find the critical points of \(f\text{,}\) we set both \(f_x=0\) and \(f_y=0\text{.}\) We first notice by inspection setting \(x=0\) satisfies \(f_y=0\) automatically, and that \(f_x (0,y) = y(y-15)\text{.}\) Hence, \((0,0)\) and \((0,15)\) must be critical points. Additionally, setting \(y=0\) automatically satisfies \(f_z = 0\text{,}\) and \(f_y(x,0) = 5x(x-3)\text{.}\) Thereofore, \((3,0)\) is another critical point. Now assume that \(x, y \neq 0\text{:}\)

   \begin{equation*} f_x = 0 \implies y= 15-10x\text{.} \end{equation*}

   Therefore, we need

   \begin{equation*} 5x + 2 (15-10x) - 15 = 0 \implies x = 1\text{.} \end{equation*}

   When \(x=1\text{,}\) we must have \(y=15-10(1) = 5\text{.}\) Hence, \(f\) has four critical points at \((0,0)\text{,}\) \((0,15)\text{,}\) \((3,0)\) and \((1,5)\text{.}\) To classify these points, we construct the discriminant,

   \begin{equation*} D(x,y) = f_{xx}f_{yy}-\left(f_{xy}\right)^2 = (10y)(2x)-(10x+2y-15)^2\text{.} \end{equation*}

   We now classify the four critical points:

   Crit. pt \((x,y)\)	\(D(x,y)\)	\(f_{xx}(x,y)\)	Conclusion
   \((0,0)\)	\(\lt 0\)	-	saddle
   \((0,15)\)	\(\lt 0\)	-	saddle
   \((3,0)\)	\(\lt 0\)	-	saddle
   \((1,5)\)	\(>0\)	\(>0\)	rel. min

8. \(\displaystyle f(x,y)=x^2y-xy^2+4xy-4x^2-4y^2\)

   Answer
   maximum at \((0,0)\)
   Solution

   The first-order partial derivatives of \(f\) are

   \begin{equation*} f_x = 2xy-y^2+4y-8x, \ \ \text{ and } \ \ f_y = x^2-2xy + 4x-8y\text{.} \end{equation*}

   To find the critical points of \(f\text{,}\) we set both \(f_x=0\) and \(f_y=0\text{.}\) By inspection, we see that the point \((0,0)\) is a solution to this system of equations. To find any other critical point(s), we now assume that \(x, y \neq 0\text{:}\)

   \begin{equation*} f_x = 0 \implies (2y-8) x = y^2-4y\text{.} \end{equation*}

   If \(y=4\text{,}\) we see that we both \(x=8\) and \(x=-4\) satistfy the system of equations. Thus, \((8,4)\) and \((-4,4)\) are two additional critical points. We now assume that \(y\neq 4\text{,}\) and let \(x= y\frac{y-4}{2y-8}\text{.}\) Then the condition \(f_y = 0\) becomes:

   \begin{equation*} f_y = 0 \implies \left(\frac{y^2-4y}{2y-8}\right)^2 - 2y \frac{y^2-4y}{2y-8} + 4 \frac{y^2-4y}{2y-8} - 8y = 0\text{.} \end{equation*}

   Since \(y\neq 0\text{,}\) we can divide through by a factor of \(x\text{:}\)

   \begin{equation*} y\left(\frac{y-4}{2y-8}\right) - 2y + 4 - 8\left(\frac{2y-8}{y-4}\right) = 0 \implies y = -8\text{.} \end{equation*}

   When \(y=-8\text{,}\) we need \(x=-4\text{.}\) In total, \(f\) has 4 critical points: \((0,0)\text{,}\) \((8,4)\text{,}\) \((-4,4)\) and \((-8,-4)\text{.}\) To classify these critical points, we construct the discriminant,

   \begin{equation*} D(x,y) = f_{xx}f_{yy}-\left(f_{xy}\right)^2 = (2y-8)(-2x-8) - (2x-2y+4)^2\text{.} \end{equation*}

   We now classify these critical points:

   Crit. pt \((x,y)\)	\(D(x,y)\)	\(f_{xx}(x,y)\)	Conclusion
   \((0,0)\)	\(>0\)	\(\lt 0\)	rel. max
   \((8,4)\)	\(\lt 0\)	-	saddle
   \((-4,4)\)	\(\lt 0\)	-	saddle
   \((-8,-4)\)	\(\lt 0\)	-	saddle

9. \(\displaystyle f(x,y)=e^{-x^3/3+x-y^2}\)

   Answer
   saddle at \((-1,0)\text{,}\) maximum at \((1,0)\)
   Solution

   The first-order partial derivatives of \(f\) are

   \begin{equation*} f_x = \left(-x^2+1\right)e^{-\frac{x^3}{3} + x - y^2} \ \ \text{ and } \ \ f_y = -2y e^{-\frac{x^3}{3} + x - y^2}\text{.} \end{equation*}

   To find the critical points of \(f\text{,}\) we set both \(f_x=0\) and \(f_y=0\text{.}\) We see that

   \begin{equation*} f_x = 0 \implies x = \pm 1, \ \ \text{ and } f_y = 0 \implies y = 0\text{.} \end{equation*}

   Thus, \(f\) has two critical points: \((1, 0)\) and \((-1,0)\text{.}\) We now construct the discriminant:

   \begin{equation*} D(x,y) = f_{xx}f_{yy}-\left(f_{xy}\right)^2 = e^{-\frac{x^3}{3} + x - y^2}\bigl(2(x^4-2x^2-2x+1)(2y^2-1)-(2(x^2-1)y)\bigr)\text{.} \end{equation*}

   We now classify these critical points:

   Crit. pt \((x,y)\)	\(D(x,y)\)	\(f_{xx}(x,y)\)	Conclusion
   \((1,0)\)	\(>0\)	\(\lt 0\)	rel. max
   \((-1,0)\)	\(\lt 0\)	-	saddle

10. \(\displaystyle f(x,y)=2x^2+2xy+2y^2-6x\)

    Answer
    minimum at \((2,-1)\)
    Solution

    The first-order partial derivatives of \(f\) are

    \begin{equation*} f_x = 4x+2y-6, \ \ \text{ and } \ \ f_y = 2x+4y\text{.} \end{equation*}

    To find the critical points of \(f\text{,}\) we set both \(f_x=0\) and \(f_y=0\text{.}\) We see that

    \begin{equation*} f_y = 0 \implies x = -2y\text{,} \end{equation*}

    and so

    \begin{equation*} f_x = 0 \implies 4(-2y)+2y-6 = 0 \implies y=-1\text{.} \end{equation*}

    When \(y=-1\text{,}\) we need \(x=2\text{.}\) Thus, \(f\) has one critical point at \((2,-1)\text{.}\) To classify this critical point, we construct the discriminant:

    \begin{equation*} D(x,y) = f_{xx}f_{yy}-\left(f_{xy}\right)^2 = (4)(4) - 2 = 14 > 0\text{.} \end{equation*}

    Since the discriminant and \(f_{xx}\) are both positive everywhere on the domain of \(f\text{,}\) we conclude that \((2,-1)\) must be a relative minimum of \(f\text{.}\)

11. \(\displaystyle f(x,y)=x^4+y^4-4xy\)

    Answer
    minimum at \((-1,-1)\text{,}\) maximum at \((1,1)\text{,}\) saddle at \((0,0)\)
    Solution

    The first-order partial derivatives of \(f\) are

    \begin{equation*} f_x = 4x^3-4y \ \ \text{ and } \ \ f_y = 4y^3-4x\text{.} \end{equation*}

    To find the critical points of \(f\text{,}\) we set both \(f_x=0\) and \(f_y=0\text{.}\) We first observe that \((0,0)\) must be a critical point. We now assume that \(x,y \neq 0\text{.}\) Then:

    \begin{equation*} f_x = 0 \implies y = x^3\text{,} \end{equation*}

    and so

    \begin{equation*} f_y = 0 \implies x^9 = x\text{.} \end{equation*}

    Thus, \(x = \pm 1\text{.}\) When \(x = -1\text{,}\) we need \(y = (-1)^3 = -1\text{,}\) and when \(x=1\text{,}\) we need \(y=1^3 = 1\text{.}\) Hence, there are three critical points in total: \((0,0)\text{,}\) \((1,1)\) and \((-1,-1)\text{.}\) We now construct the discriminant:

    \begin{equation*} D(x,y) = f_{xx}f_{yy}-\left(f_{xy}\right)^2 = (12x^2)(12y^2) - 16\text{.} \end{equation*}

    Using the discriminant, we can classify these critical points:

    Crit. pt \((x,y)\)	\(D(x,y)\)	\(f_{xx}(x,y)\)	Conclusion
    \((0,0)\)	\(\lt 0\)	-	saddle
    \((1,1)\)	\(>0\)	\(>0\)	rel. min
    \((-1,-1)\)	\(>0\)	\(>0\)	rel. min

12. \(\displaystyle f(x,y)=x^2y-2xy^2+3xy+4\)

    Answer
    minimum at \((-1,1/2)\text{,}\) saddle at \((-3,0)\text{,}\) \((0,0)\) and \((0,3/2)\text{.}\)
    Solution

    The first-order partial derivatives of \(f\) are

    \begin{equation*} f_x = 2xy - 2y^2 + 3y \ \ \text{ and } \ \ f_y = x^2-4xy + 3x\text{.} \end{equation*}

    To find the critical points of \(f\text{,}\) we set both \(f_x=0\) and \(f_y=0\text{.}\) We first observe that taking \(y=0\) satisfies \(f_x = 0\) automatically. Since \(f_y(x,0) = x(x+3)\text{,}\) we see that \((0,0)\) and \((-3,0)\) must be critical points. Furthermore, setting \(x=0\) satisfies \(f_y = 0\) automcatically, and \(f_x(0,y) = y(-2y+3)\text{.}\) Therefore, \((0, 3/2)\) is another critical point. To find any remaining critical points, we assume that \(x,y \neq 0\text{.}\) Then:

    \begin{equation*} f_x = 0 \implies 2x - 2y + 3 = 0 \implies x = y - \frac{3}{2}\text{.} \end{equation*}

    Then,

    \begin{equation*} f_y = 0 \implies (y-3/2) - 4y + 3 = 0 \implies y = \frac{1}{2}\text{.} \end{equation*}

    Hence, \((-1,1/2)\) is another critical point of \(f\text{.}\) We now construct the discriminant:

    \begin{equation*} D(x,y) = f_{xx}f_{yy}-\left(f_{xy}\right)^2 = (2y)(-4x)-(2x-4y+3)^2\text{.} \end{equation*}

    Using the discriminant, we can classify these critical points:

    Crit. pt \((x,y)\)	\(D(x,y)\)	\(f_{xx}(x,y)\)	Conclusion
    \((0,0)\)	\(\lt 0\)	-	saddle
    \((-3,0)\)	\(\lt 0\)	-	saddle
    \((0,3/2)\)	\(\lt 0\)	-	saddle
    \((-1,1/2)\)	\(>0\)	\(>0\)	rel. min

Find the absolute maximum and minimum points of \(f=x^2+3y-3xy\) over the region bounded by \(y=x\text{,}\) \(y=0\text{,}\) and \(x=2\text{.}\)

The triangular region bounded by the lines \(y=x\text{,}\) \(y=0\) and \(x=2\) is shown below.

We first find the first-order partial derivatives of \(f\text{:}\)

\begin{gather*} f_x = 2x-3y\\ f_y = 3-3x \end{gather*}

We now find and classify any relative extrema. Since

\begin{equation*} f_y = 0 \implies x = 1, \ \ \text{ and } f_x = 0 \implies 2x = 3y\text{,} \end{equation*}

we see that \((1, 3/2)\) is the only point which satisfies both \(f_x = f_y= 0\text{.}\) This is located inside the bounded region, and so it is a critical point of \(f\text{.}\)

The boundary of the region is composed of three lines: \(y=0\) for \(x \in [0,2]\text{,}\) \(x=2\) for \(y \in [0,2]\) and \(y=x\) for \(x \in [0,2]\text{.}\) And so we have the following boundary points: \((0,0)\text{,}\) \((2,2)\) and \((2,0)\text{.}\)

Over the first line, we have

\begin{equation*} f(x,0) = a(x) = x^2\text{,} \end{equation*}

which gives no additional critical points.

Over the second line, we have

\begin{equation*} f(x,x) = b(x) = -2x^2+3x\text{.} \end{equation*}

Therefore, the point \((3/4, 3/4)\) is an aditional critical point.

Finally, over the last boundary line,

\begin{equation*} f(2, y) = c(y) = 4-3y\text{,} \end{equation*}

which gives no additional critical points.

We now compare the values of \(f(x,y)\) at all of the identified critical and boundary points:

	\(f(x,y)\)
\((0,0)\)	0
\((2,2)\)	-2
\((2,0)\)	4
\((3/4, 3/4)\)	1.125
\((1,3/2)\)	1

Therefore, \(f\) has an absolute maximum of 4 at \((2,0)\) and an absolute minimum of -2 at \((2,2)\text{.}\)

A six-sided rectangular box is to hold \(1/2\) cubic meter; what shape should the box be to minimize surface area?

A certain company produces and sells flat screen TVs. The total weekly revenue and cost in dollars is given by

respectively, where \(x\) denotes the number of deluxe units and \(y\) the number of standard units produced and sold each week. Find the company's maximum profit per week, and the number of deluxe and standard units that must be produced and sold to achieve this profit.

A certain company produces and sells hand-made wooden living room tables. The total monthly revenue and cost in dollars is given by

respectively, where \(x\) denotes the number of finished tables and \(y\) the number of unfinished tables produced and sold each month. Find the company's maximum profit per month, and the number of finished and unfinished tables that must be produced and sold to achieve this profit.

The post office will accept packages whose combined length and girth is at most 130 inches. (Girth is the maximum distance around the package perpendicular to the length; for a rectangular box, the length is the largest of the three dimensions.) What is the largest volume that can be sent in a rectangular box?

The bottom of a rectangular box costs twice as much per unit area as the sides and top. Find the shape for a given volume that will minimize cost.

Using the methods of this section, find the shortest distance from the origin to the plane \(x+y+z=10\text{.}\)

Using the methods of this section, find the shortest distance from the point \((x_0,y_0,z_0)\) to the plane \(ax+by+cz=d\text{.}\) You may assume that \(c\not=0\text{;}\) use of Sage or similar software is recommended.

A trough is to be formed by bending up two sides of a long metal rectangle so that the cross-section of the trough is an isosceles trapezoid. If the width of the metal sheet is 2 meters, how should it be bent to maximize the volume of the trough?

A labelled cross-section of the trough is shown below:

Then the problem is to maximize the area of this trapezoid (since this will maximize the volume of the trough for any given length) such that \(a+a+b = 2\) metres. We first find an expression for the area, which is the area of the rectangular portion plus the area of the two right traingles:

\begin{equation*} A(a,b,h)= b\times h + h \times \sqrt{a^2-h^2}\text{.} \end{equation*}

We now use the constraint to write the area as a function of two variables:

\begin{equation*} A(a,h) = h (2-2a + \sqrt{a^2-h^2}\text{.} \end{equation*}

The first-order partial derivatives of \(A\) are

\begin{equation*} A_a = \frac{ah}{\sqrt{a^2-h^2}}-2h, \ \ \text{ and } \ \ A_h = -\frac{h^2}{\sqrt{a^2 - h^2}} + \sqrt{a^2 - h^2} -2a+2\text{.} \end{equation*}

To find any critical points, we set \(A_a = A_h = 0\) (we will assume that all of \(a,h > 0\)):

\begin{equation*} A_a = 0 \implies a=2\sqrt{a^2-h^2} \implies h^2 = \frac{3}{4} a^2\text{,} \end{equation*}

and so plugging this condition into \(A_h = 0\) gives

\begin{equation*} -2a+2 + \sqrt{\frac{1}{4}a^2} - \frac{3a^2}{4\sqrt{\frac{1}{4}a^2}} = -2a+2+\frac{a}{2} - \frac{3a}{2} = 0 \implies a = \frac{2}{3}\text{.} \end{equation*}

The corresponding value of \(h\) is therefore

\begin{equation*} h = \sqrt{\frac{3}{4}\left(\frac{2}{3}\right)^2} = \frac{1}{\sqrt{3}}\text{.} \end{equation*}

Finally, we compute the corresponding value for \(b\text{:}\)

\begin{equation*} b = 2 - 2\left(\frac{2}{3}\right) = \frac{2}{3}\text{.} \end{equation*}

We conclude that the sides and bottom should be \(2/3\) m long, and the height of the trough should be \(1/\sqrt{3}\) m (this corresponds to an angle of \(\pi/3\)).

Given the three points \((1,4)\text{,}\) \((5,2)\text{,}\) and \((3,-2)\text{,}\) \(\ds(x-1)^2+(y-4)^2+(x-5)^2+(y-2)^2+(x-3)^2+(y+2)^2\) is the sum of the squares of the distances from point \((x,y)\) to the three points. Find \(x\) and \(y\) so that this quantity is minimized.

We wish to find the point \((x,y)\) which minimizes the sum of squares distance to the three given points. Let

\begin{equation*} S(x,y) = (x-1)^2 + (y-4)^2 + (x-5)^2 + (y-2)^2+(x-3)^2+(y+2)^2\text{.} \end{equation*}

Now compute the first-order partial derivatives of \(s\text{:}\)

\begin{equation*} S_x = 2(x-1)+2(x-5)+2(x-3) = 6x-18, \ \ \text{ and } \ \ S_y = 2(y-4)+2(y-2)+2(y+2)=6y-8\text{.} \end{equation*}

To find the critical point(s) of \(S\text{,}\) we set both \(S_x=0\) and \(S_y=0\text{.}\) This gives exactly one critical point at \((3, 4/3)\text{.}\)

We can use the Second Derivative Test to classify this critical point:

\begin{equation*} D(x,y) = 6(6) - 0 = 36 > 0\text{,} \end{equation*}

and \(f_{xx}(x,y) > 0\text{.}\) Hence, this critical point is a relative minimum. Since it is the only critical point, and the domain of \(S\) is unbounded, it must be the absolute minimum. We illustrate the solution below.

Suppose that \(f(x,y)=x^2+y^2+kxy\text{.}\) Find and classify the critical points, and discuss how they change when \(k\) takes on different values.

Solution

We compute the first-order partial derivatives of \(f(x,y) = x^2+y^2+kxy\text{,}\) for some \(k\in \mathbb{R}\text{:}\)

\begin{equation*} f_x = 2x+ky, \ \ \text{ and } \ \ f_y = 2y+kx\text{.} \end{equation*}

To find the critical point(s) of \(f\text{,}\) we set both \(f_x=0\) and \(f_y=0\text{:}\)

\begin{equation*} f_x = 0 \implies x = -\frac{ky}{2}\text{,} \end{equation*}

and so

\begin{equation*} f_y = 0 \implies 2y + k\left(-\frac{ky}{2}\right) = 0\text{.} \end{equation*}

Hence, the only critical point is at \((0,0)\text{.}\) To classify this critical point, we use the Second Derivative Test:

\begin{equation*} D(0,0) = 2(2) -k^2 = 4-k^2\text{,} \end{equation*}

and

\begin{equation*} f_{xx}(0,0) = 2\text{.} \end{equation*}

As a function of \(k\text{,}\) the discriminant is therefore an upside down parabola:

We therefore have three cases:

1. For \(k \in (-2,2)\text{,}\) we have \(D(0,0) > 0\) and \(f_{xx}(0,0) > 0\) and so the point \((0,0)\) is a relative minimum.

2. For \(k\in(-\infty,-2)\cup(2,\infty)\text{,}\) then \(D(0,0)\lt 0\) and so the point \((0,0)\) is a saddle point.

3. For \(k=-2,2\text{,}\) then \(D(0,0)=0\) and the Second Derivative Test is inconclusive. Further investigation of the surfaces \(z=x^2+y^2-2xy\) and \(z=x^2+y^2+2xy\) reveals that \((0,0)\) is a relative minimum in both cases.

Find the shortest distance from the point \((0,b)\) to the parabola \(y=x^2\text{.}\)

Find the shortest distance from the point \((0,0,b)\) to the paraboloid \(z=x^2+y^2\text{.}\)

Consider the function \(f(x,y)=x^3-3x^2y+y^3\text{.}\)

1. Show that \((0,0)\) is the only critical point of \(f\text{.}\)

   Solution

   The first partial derivatives of \(f\) are

   \begin{equation*} f_x = \diffp{}{x} \left(x^3-3x^2y+y^3\right) = 3x^2-6xy\text{,} \end{equation*}

   and

   \begin{equation*} f_y=\diffp{}{y}\left(x^3-3x^2y+y^3\right) = -3x^2+3y^2\text{.} \end{equation*}

   To find the critical points of \(f\text{,}\) we need to solve the system of equations

   \begin{equation*} f_x = 3x^2-6xy = 0, \ \ \text{ and } \ \ f_y= -3x^2 + 3y^2 =0\text{.} \end{equation*}

   From the second equation, let

   \begin{equation*} 3x^2=3y^2 \implies y = \pm x\text{.} \end{equation*}

   Now substitute these values of \(y\) into the first equation,

   \begin{equation*} \begin{array}{ccc} 3x^2-6x(-x)= 0 \amp \text{ or } \amp 3x^2-6x(x)=0\\ 8x^2 = 0 \amp \amp -3x^2=0 \end{array} \end{equation*}

   Therefore, \(x=0\) is the only solution, which implies that \(y=0\text{.}\) Thus, \((0,0)\) is our only critical point.

2. Show that the Discriminant Test is inconclusive for \(f\text{.}\)

   Solution

   We first find the second partial derivatives of \(f\text{.}\)

   \begin{equation*} f_{xx} = \diffp{}{x} \left(3x^2-6xy\right) = 6x-6y \end{equation*}

   \begin{equation*} f_{yy} = \diffp{}{y} \left(-3x^2+ 3y^2\right)= 6y \end{equation*}

   \begin{equation*} f_{xy} = \diffp{}{y} \left(3x^2-6xy\right)=-6x = f_{yx} \end{equation*}

   Using the formula for the discriminant, we get

   \begin{equation*} D=f_{xx}f_{yy}-f_{xy}^2 = 36y\left(x-y\right)+36x^2\text{.} \end{equation*}

   So at the critical point \((0,0)\text{,}\) \(D=0\) and the test is inconclusive.

3. Determine the cross-sections of \(f\) obtained by setting \(y=kx\) for various values of \(k\text{.}\) Solution

   We show cross-sections of \(f\) where \(y=kx\) for \(k=-2,0,2\) below.

   

4. What kind of critical point is \((0,0)\text{?}\)

   Solution

   The above cross-sections indicate that the critical point \((0,0,0)\) is a saddle point of \(f\text{.}\) We justify this statement with the following surface plot:

   

Find the volume of the largest rectangular box with edges parallel to the axes that can be inscribed in the ellipsoid \(2x^2+72y^2+18z^2=288\text{.}\)