Calculus Optimization Problem:
What are the dimensions of the poster with the smallest total area?

A rectangular 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 when 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. $\quad \cmark$

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