solve each optimization problem. use the provided box to sketch the problem setup. 1) a graphic designer is…

Explanation:

Step1: Let the dimensions of the photo

Let the length of the photo be $x$ inches and the width be $y$ inches. We know that the area of the photo is $xy = 98$, so $y=\frac{98}{x}$.

Step2: Find the dimensions of the poster

The length of the poster $L=x + 2$ (1 - inch border on each side) and the width of the poster $W=y + 4$ (2 - inch border on top and bottom). Substitute $y=\frac{98}{x}$ into the width formula, we get $W=\frac{98}{x}+4$.

Step3: Find the area function of the poster

The area of the poster $A=(x + 2)(\frac{98}{x}+4)=98+4x+\frac{196}{x}+8=106 + 4x+\frac{196}{x}$, where $x>0$.

Step4: Take the derivative of the area function

Using the power - rule, if $A(x)=106 + 4x+\frac{196}{x}=106 + 4x+196x^{-1}$, then $A^\prime(x)=4-196x^{-2}=4-\frac{196}{x^{2}}$.

Step5: Set the derivative equal to zero

Set $A^\prime(x)=0$, so $4-\frac{196}{x^{2}} = 0$. Then $4=\frac{196}{x^{2}}$, and $x^{2}=\frac{196}{4}=49$. Solving for $x$, we get $x = 7$ (we ignore $x=-7$ since $x>0$).

Step6: Find the second - derivative

$A^{\prime\prime}(x)=\frac{392}{x^{3}}$. When $x = 7$, $A^{\prime\prime}(7)=\frac{392}{7^{3}}=\frac{392}{343}>0$, so the area function has a minimum at $x = 7$.

Step7: Find the dimensions of the poster

When $x = 7$, $y=\frac{98}{7}=14$. The length of the poster $L=x + 2=7 + 2=9$ inches and the width of the poster $W=y + 4=14 + 4=18$ inches.

Answer:

The dimensions of the poster should be 9 inches by 18 inches.