Optimization problems

Example 1: A farmer has $2400$ ft of fencing. The farmer wants to fence a rectangular field next to a straight river. The side along the river needs no fence. What size gives the largest area?

Solution:

Let $x$ be the length of each side that touches the river, and $y$ the length of the side opposite the river.

Formula: The area is $A=xy$.

Constraint: The fence covers three sides:

$2x+y=2400$,   so   $y=2400-2x$.

One variable:

$\begin{align*}&A(x)\\&=x(2400-2x)\\&=2400x-2x^2\end{align*}$

Domain: Both lengths must be at least $0$. So

$0\le x\le 1200$.

Maximize: $A'(x)=2400-4x$. It is $0$ at $x=600$.

• $A(0)=0$
• $A(600)=600\cdot 1200=720{,}000$
• $A(1200)=0$

Answer: The sides touching the river should be $600$ ft long, and the side opposite the river $1200$ ft long. The largest area is $720{,}000$ ft$^2$.

Example 2: An open box is made from a square sheet of cardboard, $12$ inches on each side. Equal squares are cut from the corners, and the sides are folded up. What size of square gives the largest volume?

Solution:

Let $x$ be the side of each cut square, in inches. Then the box is $x$ tall, and its base is a square with side $12-2x$.

Formula:

$V(x)=x(12-2x)^2$

Domain: $0\le x\le 6$. (At $x=6$, nothing is left of the base.)

Maximize: First, by the chain rule, the derivative of $(12-2x)^2$ is

$2(12-2x)(-2)=-4(12-2x)$.

Now use the product rule, and factor out $(12-2x)$:

$\begin{align*}&V'(x)\\&=(12-2x)^2-4x(12-2x)\\&=(12-2x)\big[(12-2x)-4x\big]\\&=(12-2x)(12-6x)\end{align*}$

So $V'(x)=0$ at $x=6$ or $x=2$.

• $V(0)=0$
• $V(2)=2\cdot 8^2=128$
• $V(6)=0$

Answer: Cut $2$-inch squares from the corners. The box is $8$ in by $8$ in by $2$ in, and its volume is $128$ in$^3$.

Example 3: A can in the shape of a cylinder must hold $1000$ cm$^3$ (one liter). What radius and height use the least metal?

Solution:

Let $r$ be the radius and $h$ the height, in centimeters. The metal is the surface area: the top, the bottom, and the side.

Formula:

$A=2\pi r^2+2\pi rh$

(The top and bottom are circles of area $\pi r^2$. The side unrolls into a rectangle, $2\pi r$ wide and $h$ tall.)

Constraint: The volume is $1000$:

$\pi r^2h=1000$,   so   $h=\dfrac{1000}{\pi r^2}$.

One variable:

$\begin{align*}&A(r)\\&=2\pi r^2+2\pi r\cdot\dfrac{1000}{\pi r^2}\\&=2\pi r^2+\dfrac{2000}{r}\end{align*}$

Domain: $r>0$. This is an open interval, so use the first derivative test.

Minimize:

$A'(r)=4\pi r-\dfrac{2000}{r^2}$

Set it equal to $0$:

$\begin{align*}4\pi r&=\dfrac{2000}{r^2}\\r^3&=\dfrac{500}{\pi}\\r&=\sqrt[3]{\dfrac{500}{\pi}}\approx 5.42\end{align*}$

For smaller $r$, $A'(r)<0$. For larger $r$, $A'(r)>0$. So this is the absolute minimum.

The height:

$h=\dfrac{1000}{\pi r^2}\approx 10.84$

Answer: The radius should be about $5.42$ cm and the height about $10.84$ cm.

Notice that $h=2r$: the best can is exactly as tall as it is wide. (Real cans are often taller, for other reasons, such as how easy they are to hold.)

Example 4: Find the points on the parabola $y=x^2$ that are closest to the point $(0,2)$.

Solution:

The distance from a point $(x,y)$ on the parabola to $(0,2)$ is

$d=\sqrt{x^2+(y-2)^2}$.

A square root gets smaller exactly when the number inside it gets smaller. So it is easier to minimize the square of the distance:

$D=d^2=x^2+(y-2)^2$

One variable: On the parabola, $x^2=y$. So

$\begin{align*}&D(y)\\&=y+(y-2)^2\\&=y^2-3y+4\end{align*}$

Domain: $y\ge 0$, because $y=x^2$.

Minimize: $D'(y)=2y-3$. It is $0$ when

$y=\dfrac{3}{2}$.

• For $y<\tfrac{3}{2}$: $D'(y)<0$.
• For $y>\tfrac{3}{2}$: $D'(y)>0$.

So this is the absolute minimum.

Then $x^2=\dfrac{3}{2}$, so $x=\pm\sqrt{\dfrac{3}{2}}\approx\pm 1.22$.

Answer: The closest points are

$\left(\pm\sqrt{\tfrac{3}{2}},\ \tfrac{3}{2}\right)\approx(\pm 1.22,\ 1.5)$.

The smallest distance is

$d=\sqrt{\tfrac{3}{2}+\tfrac{1}{4}}=\dfrac{\sqrt{7}}{2}\approx 1.32$.