Chapter 4.4, Problem 13E

To determine

To find: The overall dimensions that will minimize the amount of paper used.

Answer to Problem 13E

The required dimension is $18\text{ in}\text{. by 9 in}\text{.}$

Explanation of Solution

Given information: A rectangular poster that contains $50\text{ in}{\text{.}}^2$ of printing with $4\text{ in}\text{.}$ margin at the top and bottom, and a $2\text{ in}\text{.}$ margin on each side.

Concept used: The minimum value of the function $f\left(x\right)$ will be at the critical point of that function.

Calculation:

Consider the height of the printed area as $x$ .

So, according to the given statement, the width of the printed area will be $\frac{50}{x}$ .

It is given that a rectangular poster contains $50\text{ in}{\text{.}}^2$ of printing with $4\text{ in}\text{.}$ margin at the top and bottom, and a $2\text{ in}\text{.}$ margin on each side.

So, the overall dimensions will be $\left(x+8\right)$ by $\left(\frac{50}{x}+4\right)$ .

The product of $\left(x+8\right)$ and $\left(\frac{50}{x}+4\right)$ will be the amount of the paper used, that is, $A$ .

  $\begin{array}{c}A=\left(x+8\right)\left(\frac{50}{x}+4\right)\\ =4x+\frac{400}{x}+82\end{array}$

Differentiate the obtained function with respect to $x$ .

  $\frac{dA}{dx}=4-400x^{-2}$

Equate the obtained derivative to 0 and solve for $x$ .

  $\begin{array}{c}\frac{dA}{dx}=0\\ 4-400x^{-2}=0\\ x=\pm 10\end{array}$

Since the height can never be negative, the value of $x$ is 10 feet.

Now, differentiate the equation, $\frac{dA}{dx}=4-400x^{-2}$ , with respect to $x$ .

  $\frac{d^{2}A}{dx}=\frac{800}{x^3}$

Now, substitute 10 for $x$ into the obtained second derivative and simplify it.

  $\begin{array}{c}{\left.\frac{d^{2}A}{dx}\right|}_{x=10}=\frac{800}{{10}^3}\\ =\frac{4}{5}\end{array}$

Since at $x=10$ , $\frac{d^{2}A}{d{x}^2}>0$ , the dimensions of the paper will be minimum at $x=10$ .

Substitute $x=10$ into the expressions $\left(x+8\right)$ and $\left(\frac{50}{x}+4\right)$ , and simplify it.

  $\left(10+8\right)=18$

  $\left(\frac{50}{10}+4\right)=9$

Therefore, the required dimensions is $18\text{ in}\text{. by 9 in}\text{.}$