## Optimization

### Calculus Optimization Problem:

What are the dimensions of the poster with the smallest total area?

A poster must have a printed area of 320 cm$^2$. It will have top and bottom margins that are 5 cm each, and side margins that are 4 cm. What are the dimensions of the poster with the smallest total area?

### Calculus Solution

We’ll use our standard Optimization Problem Solving Strategy to develop our solution. (Link will open in a new tab.)

*Stage I: Develop the function*.

Your first job is to develop a function that represents the quantity you want to optimize. It can depend on only *one* variable. The steps:

**1. Draw a picture of the physical situation.**

See the figure. We’ve called the width of the printed area *x*, and its length *y*. We can then write the printed area as

$$A_{\text{printed}} =xy = 320 \text{ cm}^2$$

Note that this picture captures the key features of the situation, and we wouldn’t make good progress without it!

**2. Write an equation that relates the quantity you want to optimize in terms of the relevant variables.**

We see from the figure that the poster has total width $w = x + 8$ cm, and total length $l = y + 10$ cm. Its total area is thus

$$A_{\text{total}} = (x+8)(y+10)$$

**3. If necessary, use other given information to rewrite your equation in terms of a single variable.**

The poster’s area $A_{\text{total}}$ currently depends on two variables, *x* and *y*. In order to proceed, we must use other information we’re given to rewrite the area in terms of just *one* of those variables. Let’s choose the width *x* as that single variable.

We must then eliminate the height *y* as a variable. To do so, recall that the printed area is given by

$$A_{\text{printed}} =xy = 320 \text{ cm}^2$$

Hence we can write *y* in terms of *x* as

$$y = \dfrac{320}{x}$$

Substituting this expression for *y* into our expression for the poster’s total area $A_{\text{total}}$ gives:

\begin{align*}

A_{\text{total}} &= (x+8)(y+10)\\[8px]
&= (x+8)\left( \dfrac{320}{x}+10 \right) \\[8px]
&= 10 (x+8)\left( \dfrac{32}{x}+1 \right) \\[8px]
&= 10\left( 32 + x + \frac{256}{x} + 8 \right) \\[8px]
&= 10 \left( x + \frac{256}{x} + 40 \right)

\end{align*}

We’ve graphed the function, a step you probably wouldn’t do yourself — but we want to emphasize that everything you’ve done so far is to create a function that you’re now going to minimize. One choice would be to closely examine the graph to determine the value of *x* that minimizes *A* . . . but instead we’re going to use the max/min techniques you learned recently!

*Stage II: Maximize or minimize the function.*

You now have a standard max/min problem to solve.

**4. Take the derivative of your equation with respect to your single variable. Then find the critical points.**

\begin{align*}

\frac{dA_{\text{total}}}{dx} &= \frac{d}{dx} \left[ 10 \left( x + \frac{256}{x} + 40 \right) \right] \\[8px]
&= 10 \frac{d}{dx} \left[ \left( x + \frac{256}{x} + 40 \right) \right] \\[8px]
&= 10 \left[ 1 – \frac{256}{x^2} \right]
\end{align*}

The critical point occurs when $\dfrac{dA_{\text{total}}}{dx} = 0$:

\begin{align*}

\frac{dA_{\text{total}}}{dx} = 0 &= 10 \left[ 1 – \frac{256}{x^2} \right] \\[8px]
0 &= 1 – \frac{256}{x^2} \\[8px]
x^2 &= 256 \\[8px]
x &= \sqrt{256} = 16 \text{ cm}\\

\end{align*}

**5. Determine the maxima and minima as necessary. Remember to check the endpoints if there are any.**

Recall that we found above that $y = \dfrac{320}{x}$. Hence hen the printed area’s width is $x = 16$ cm, its height is

$$ y = \dfrac{320}{16} = 20 \text{ cm}$$

**6. Justify your maxima or minima either by reasoning about the physical situation, or with the first derivative test, or with the second derivative test.**

Let’s use the second derivative test to verify that the critical point we’ve found represents a minimum. We found above that the first derivative is

$$\frac{dA_{\text{total}}}{dx} = 10 \left( 1 – \frac{256}{x^2} \right)$$

The second derivative is thus

\[ \begin{align*}

\dfrac{d^2A_{\text{total}}}{dx^2} &= 10\dfrac{d}{dx}(1) – 10(256)\dfrac{d}{dx}\left(\frac{1}{x^2} \right) \\[8px]
&= 0 – (-2)(10)(256)\dfrac{1}{x^3} \\[8px]
&= \dfrac{(2)(10)(256)}{x^3}

\end{align*} \]
Since the width is always positive ($x > 0$), this second derivative is always positive, and so our single critical point gives us a *minimum* value for the poster’s area.

**7. Finally, check to make sure you have answered the question as asked: x or y values, or coordinates, or a maximum area, or a shortest time, or . . . .**

*Careful!*We’ve found

*x*and

*y*for the poster’s printed area, but the question asks for the dimensions of the overall poster. Those values are

\begin{align*}

\text{width} &= x + 8 = 16 + 8 = 24 \text{ cm} \\[8px] \text{length} &= y + 10 = 20 + 10 = 30 \text{ cm}

\end{align*}

The minimum dimensions of the poster are thus width $x = 24$ cm and length $y = 30$ cm. $\quad \cmark$

Return to Optimization Problems

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