Find the critical points of $f(x,y)=4+x^3+y^3-3xy$.

Solution

Find the first derivatives: $\begin{aligned} f_x (x,y) &= 3x^2-3y\ f_y (x,y) &= 3y^2-3x \end{aligned}$

Since these exist everywhere, we simply need to find where they equal zero: $3x^2-3y = 0 \longrightarrow y=x^2$

If $y=x^2$, then $\begin{aligned} 3y^2-3x &= 0\ 3x^4-3x &= 0\ 3x(x^3-1) &= 0\ \implies x = 0,1 \end{aligned}$

Since $y=x^2$, these $x$ values correspond to $y$ values of $0$ and $1$, respectively. Thus the critical points are $\ans{(0,0) \textrm{ and } (1,1).}$

Find the local maxima and minima of $f(x,y)=4+x^3+y^3-3xy$.

Solution

We already found the critical points: $(0,0) \textrm{ and } (1,1)$

Next, find the discriminant and evaluate it at the critical points: $\begin{aligned} D (x,y) &= f_{xx} (x,y)\ f_{yy} (x,y) - \left(f_{xy} (x,y)\right)^2\ &= 36xy-9\ \ D (0,0) &= -9 \longrightarrow \textrm{ Saddle Point}\ D (1,1) &= 27 \textrm{ and } f_{xx} (1,1) = 6 > 0 \longrightarrow \textrm{ Minimum} \end{aligned}$

We conclude that there is a saddle point at $(0,0)$ and a local minimum at $(1,1)$.

Find the local maxima and minima of $f(x,y)=3x^2y+y^3-3x^2-3y^2+2$.

Solution

First, find the critical points. Note that the first partial derivatives exist everywhere, so we need to find the points where $f_x$ and $f_y$ are zero. $\begin{aligned} f_x &= 6xy-6x = 0 \longrightarrow x(y-1)=0 \longrightarrow x = 0 \textrm{ or } y=1\ f_y &= 3x^2+3y^2-6y = 0 \end{aligned}$

• If $x=0$: $f_y = 3y^2-6y = 0 \longrightarrow 3y(y-2) = 0 \longrightarrow y = 0,2$

• If $y=1$: $f_y = 3x^2-3 = 0 \longrightarrow 3(x+1)(x-1) = 0 \longrightarrow x=1,-1$

Therefore, the critical points are $(0,0)$, $(0,2)$, $(1,1)$, and $(−1,1)$.

Next, find the discriminant and evaluate it at the critical points: $\begin{aligned} D (x,y) &= (6y-6)^2 - (6x)^2 = 36y^2-72y+36-36x^2\ \ D (0,0) &= 36 > 0 \textrm{ and } f_{xx} (0,0) = -6 \longrightarrow \textrm{MAX}\ D (0,2) &= 36 > 0 \textrm{ and } f_{xx} (0,2) = 6 \longrightarrow \textrm{MIN}\ D (1,1) &= -36 \longrightarrow \textrm{Saddle Point}\ D (-1,1) &= -3 \longrightarrow \textrm{Saddle Point} \end{aligned}$

Let $x$, $y$, and $z$ be nonnegative numbers that add up to $200$: $x+y+z=200$

Find values of $x$, $y$, and $z$ that minimize $x^2+y^2+z^2.$

Solution

Use the condition to write the quantity to be minimized as a function of two variables: $z=200-x-y \longrightarrow f(x,y) = x^2+y^2+(200-x-y)^2$

Next, take the first partial derivatives to find the critical points: $\begin{aligned} f_x &= 2x-2(200-x-y) = 4x+2y-400\ f_y &= 2y-2(200-x-y) = 4y+2x-400 \end{aligned}$

If $4x+2y−400=0$, $y=200−2x$. Plug this into $f_y$: $4(200-2x)+2x-400 = -6x+400 = 0 \longrightarrow x = \dfrac{200}{3}$

Using this, $y=200-2x = \dfrac{200}{3}$. Therefore, the one critical point of this function is $\left(\dfrac{200}{3},\dfrac{200}{3}\right)$.

Find the discriminant at this point: $D (x,y) = (4)(4) - (2)^2 = 12 > 0 \textrm{ and } f_{xx} = 4 > 0$

Therefore, this quantity is minimized when $\ans{x=\dfrac{200}{3},\ y=\dfrac{200}{3},\ \textrm{ and } z=\dfrac{200}{3}.}$

A shipping company handles rectangular boxes if the sum of the length, width, and height of the box does not exceed $96$ inches. Find the dimensions of the box that meets this condition and has the largest volume.

Solution

The condition in this problem is $x+y+z=96$. The volume, the quantity that we'd like to maximize, is $V=xyz$. To write this as a function of two variables, we'll use the condition to eliminate $z$: $V(x,y)=xy(96-x-y) = 96xy-x^2y-xy^2$

Take the first partial derivatives to find the critical points: $\begin{aligned} V_x &= 96y-2xy-y^2 = y(96-2x-y)\ V_y &= 96x-x^2-2xy = x(96-x-2y) \end{aligned}$

If $y(96−2x−y)=0$, then either $y=0$ or $y=96−2x$. Plug each of these into $V_y$:

• If $y=0$: $x(96-x) = 0 \longrightarrow x = 0,96$ This leads to two critical points: $(0,0)$ and $(96,0)$.

• If $y=96−2x$: $x(-96+3x) = 0 \longrightarrow x = 0, 32$ This leads to two critical points: $(0,96)$ and $(32,32)$.

We have four critical points, but we can save ourselves some work by noticing that only with the fourth one $[(32,32)]$ is the volume nonzero. We can test to verify that this is a local maximum by finding the discriminant and evaluating it at $(32,32)$:

$\begin{aligned} V_{xx} &= -2y, V_{yy} = -2x, V_{xy} = 96-2x-2y\ \ D (x,y) &= (-2y)(-2x)-(96-2x-2y)^2\ D (32,32) &= 3072 > 0 \textrm{ and } V_{xx} (32,32) = -6 < 0 \end{aligned}$

Therefore, the volume is maximized at $\ans{x=32,\ y=32,\ \textrm{ and } z=32.}$

Build an aquarium with a volume of $200$ cubic feet; if the base (made of slate) costs $5$ times as much as the sides (made of glass), find the dimensions that minimize the cost of the aquarium.

Solution

The condition in this problem is $xyz=200$. The cost, the quantity that we'd like to minimize, is $C=5xy+2xz+2yz$. To write this as a function of two variables, we'll use the condition to eliminate $z$: $C(x,y)=5xy+\dfrac{400}{y}+\dfrac{400}{x}$

Take the first partial derivatives to find the critical points: $\begin{aligned} C_x &= 5y-\dfrac{400}{x^2} = 0 \longrightarrow y=\dfrac{80}{x^2}\ C_y &= 5x-\dfrac{400}{y^2} = 0\ &= 5x-\dfrac{400}{(80/x^2)^2} = 0 \longrightarrow x^3 = 80 \longrightarrow x \approx 4.31 \end{aligned}$

We can test to verify that this is a local minimum by finding the discriminant and evaluating it at $(4.31,4.31)$: $\begin{aligned} C_{xx} &= \dfrac{800}{x^3}, C_{yy} = \dfrac{800}{y^3}, C_{xy} = 5\ \ D (x,y) &= \dfrac{640,000}{x^3y^3}-(5)^2\ D (4.31,4.31) &= 75 > 0 \textrm{ and } C_{xx} (4.31,4.31) = 10 > 0 \end{aligned}$

Therefore, the cost is minimized at $\ans{x=4.31\ ft,\ y=4.31\ ft,\ \textrm{ and } z=10.77\ ft.}$

Suppose three houses are located at $(0,0)$, $(2,0)$, and $(1,1)$. A power substation must be located at a point such that the sum of the squares of the distances between the houses and the substation is minimized.

Solution

The sum of the squares of the distances is $\begin{aligned} D &= x^2+y^2+(x-2)^2+y^2+(x-1)^2+(y-1)^2\ &= 3x^2+3y^2-6x-2y+6 \end{aligned}$

Take the first partial derivatives to find the critical points: $\begin{aligned} D_x &= 6x-6 = 0 \longrightarrow x=1\ D_y &= 6y-2 = 0 \longrightarrow y = \dfrac{1}{3}\ \end{aligned}$

We can test to verify that this is a local minimum by finding the discriminant and evaluating it at $\left(1,\dfrac{1}{3}\right)$: $\begin{aligned} D_{xx} &= 6, D_{yy} = 6, D_{xy} = 0\ \ D (x,y) &= 36 > 0 \textrm{ and } D_{xx} > 0 \end{aligned}$

Therefore, the substation should be positioned at $\ans{\left(1,\dfrac{1}{3}\right).}$