Applied Optimization Problems: Learn It 2

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Applied Optimization Problems: Learn It 2

Solving Optimization Problems

Solving Optimization Problems when the Interval Is Not Closed or Is Unbounded

In the previous examples, we considered functions on closed, bounded domains. Consequently, by the extreme value theorem, we were guaranteed that the functions had absolute extrema. Let’s now consider functions for which the domain is neither closed nor bounded.

Many functions still have at least one absolute extrema, even if the domain is not closed or the domain is unbounded. For example, the function $f(x)=x^2+4$ over $(−\infty ,\infty )$ has an absolute minimum of 4 at $x=0$ . Therefore, we can still consider functions over unbounded domains or open intervals and determine whether they have any absolute extrema.

In the following example, we look at constructing a box of least surface area with a prescribed volume. It is not difficult to show that for a closed-top box, by symmetry, among all boxes with a specified volume, a cube will have the smallest surface area. Consequently, we consider the modified problem of determining which open-topped box with a specified volume has the smallest surface area.

A rectangular box with a square base, an open top, and a volume of $216 \, \text{in}^3$ is to be constructed. What should the dimensions of the box be to minimize the surface area of the box? What is the minimum surface area?

Step 1: Draw a rectangular box and introduce the variable $x$ to represent the length of each side of the square base; let $y$ represent the height of the box. Let $S$ denote the surface area of the open-top box.

A box with square base is shown. The base has side length x, and the height is y. — Figure 8. We want to minimize the surface area of a square-based box with a given volume.

Step 2: We need to minimize the surface area. Therefore, we need to minimize $S$ .

Step 3: Since the box has an open top, we need only determine the area of the four vertical sides and the base. The area of each of the four vertical sides is $x \cdot y$ . The area of the base is $x^2$ . Therefore, the surface area of the box is

S = 4 x y + x^{2}

$S=4xy+x^2$ .

Step 4: Since the volume of this box is $x^2 y$ and the volume is given as $216 \, \text{in}^3$ , the constraint equation is

x^{2} y = 216

$x^2 y=216$ .

Solving the constraint equation for $y$ , we have $y=\frac{216}{x^2}$ . Therefore, we can write the surface area as a function of $x$ only:

S (x) = 4 x (\frac{216}{x^{2}}) + x^{2}

$S(x)=4x(\frac{216}{x^2})+x^2$ .

Therefore, $S(x)=\frac{864}{x}+x^2$ .

Step 5: Since we are requiring that $x^2 y=216$ , we cannot have $x=0$ . Therefore, we need $x>0$ . On the other hand, $x$ is allowed to have any positive value.

Note that as $x$ becomes large, the height of the box $y$ becomes correspondingly small so that $x^2 y=216$ . Similarly, as $x$ becomes small, the height of the box becomes correspondingly large.

We conclude that the domain is the open, unbounded interval $(0,\infty )$ . Note that, unlike the previous examples, we cannot reduce our problem to looking for an absolute maximum or absolute minimum over a closed, bounded interval.

However, in the next step, we discover why this function must have an absolute minimum over the interval $(0,\infty )$ .

Step 6: Note that as $x\to 0^+$ , $S(x)\to \infty$ . Also, as $x\to \infty$ , $S(x)\to \infty$ . Since $S$ is a continuous function that approaches infinity at the ends, it must have an absolute minimum at some $x\in (0,\infty )$ . This minimum must occur at a critical point of $S$ . The derivative is

S^{'} (x) = - \frac{864}{x^{2}} + 2 x

$S^{\prime}(x)=-\frac{864}{x^2}+2x$ .

Therefore, $S^{\prime}(x)=0$ when $2x=\frac{864}{x^2}$ .

Solving this equation for $x$ , we obtain $x^3=432$ , so $x=\sqrt[3]{432}=6\sqrt[3]{2}$ .

Since this is the only critical point of $S$ , the absolute minimum must occur at $x=6\sqrt[3]{2}$ (see Figure 9). When $x=6\sqrt[3]{2}$ , $y=\frac{216}{(6\sqrt[3]{2})^2}=3\sqrt[3]{2}$ in.

Therefore, the dimensions of the box should be $x=6\sqrt[3]{2}$ in and $y=3\sqrt[3]{2}$ in. With these dimensions, the surface area is:

S (6 \sqrt[3]{2}) = \frac{864}{6 \sqrt[3]{2}} + (6 \sqrt[3]{2})^{2} = 108 \sqrt[3]{4} {in}^{2}

$S(6\sqrt[3]{2})=\frac{864}{6\sqrt[3]{2}}+(6\sqrt[3]{2})^2=108\sqrt[3]{4} \, \text{in}^2$

The function S(x) = 864/x + x2 is graphed. At its minimum there is a dashed line and text that reads “Minimum surface area is 108 times the cube root of 4 square inches when x = 6 times the cube root of 2 inches.” — Figure 9. We can use a graph to determine the dimensions of a box of given the volume and the minimum surface area.

Watch the following video to see the worked solution to the example above.

Consider the same open-top box, which is to have volume $216 \, \text{in}^3$ . Suppose the cost of the material for the base is $$0.20 / \text{in}^2$ and the cost of the material for the sides is $$0.30 / \text{in}^2$ and we are trying to minimize the cost of this box. Write the cost as a function of the side lengths of the base. (Let $x$ be the side length of the base and $y$ be the height of the box.)

If the cost of one of the sides is $$0.30 / \text{in}^2$ , the cost of that side is $0.30xy$ .

$c(x)=\frac{259.2}{x}+0.2x^2$ dollars