Question

a rectangular page is to contain 12 square inches of print. the page has to have a 1 - inch margin on top and at the bottom and a 3 - inch margin on each side (see figure). find the dimensions of the page that minimize the amount of paper used. the dimensions that minimize the amount of paper used are (simplify your answers. use a comma to separate answers.)

Explanation:

Step1: Define variables

Let the length of the printed - area be $x$ inches and the width of the printed - area be $y$ inches. Then the area of the printed area is $A_{print}=xy = 12$, so $y=\frac{12}{x}$. The length of the paper $L=x + 2$ (1 - inch margin on top and bottom), and the width of the paper $W=y + 6$ (3 - inch margin on each side). The area of the paper $A=(x + 2)(y + 6)$.

Step2: Substitute $y$ into the area formula

Substitute $y=\frac{12}{x}$ into the area formula:
\[

$$\begin{align*} A&=(x + 2)(\frac{12}{x}+6)\\ &=x\cdot\frac{12}{x}+6x+2\cdot\frac{12}{x}+2\times6\\ &=12 + 6x+\frac{24}{x}+12\\ &=24 + 6x+\frac{24}{x} \end{align*}$$

\]

Step3: Find the derivative of $A$

Differentiate $A(x)=24 + 6x+\frac{24}{x}$ with respect to $x$. Using the power rule, if $y = ax^n$, then $y^\prime=anx^{n - 1}$.
\[
A^\prime(x)=6-\frac{24}{x^{2}}
\]

Step4: Set the derivative equal to zero

Set $A^\prime(x)=0$ to find the critical points:
\[

$$\begin{align*} 6-\frac{24}{x^{2}}&=0\\ 6&=\frac{24}{x^{2}}\\ x^{2}& = 4\\ x&=2 \end{align*}$$

\]
We take the positive value of $x$ since $x$ represents a length.

Step5: Find the second - derivative of $A$

Differentiate $A^\prime(x)=6-\frac{24}{x^{2}}$ with respect to $x$ to get the second - derivative:
\[
A^{\prime\prime}(x)=\frac{48}{x^{3}}
\]
When $x = 2$, $A^{\prime\prime}(2)=\frac{48}{2^{3}}=\frac{48}{8}=6>0$. So $x = 2$ gives a minimum of the area function.

Step6: Find the value of $y$

Since $y=\frac{12}{x}$ and $x = 2$, then $y = 6$.

Step7: Find the dimensions of the paper

The length of the paper $L=x + 2=2 + 2=4$ inches, and the width of the paper $W=y + 6=6+6 = 12$ inches.

Answer:

$4,12$