Answer to Question #242075 in Calculus for Carms

Question #242075

A poster must have 32 square inches of printed matter with margins of 4 inches at the top, and 2 inches at each side. Find the dimensions of the whole poster if its area is MAXIMUM.


1
Expert's answer
2021-09-28T09:03:43-0400

Let "x=" the length of printed matter, "y="the width of printed matter.

Given "xy=32."

The length of a poster will be "x+2(2)," the width of a poster will be "y+4."

The area of a poster is "A=(x+4)(y+4)."


"xy=32=>y=\\dfrac{32}{x}"

Substitute


"A=A(x)=(x+4)(\\dfrac{32}{x}+4), x>0"

"A(x)=32+4x+\\dfrac{128}{x}+16"

Find the derivative with respect to "x"


"A'(x)=(48+4x+\\dfrac{128}{x})'=4-\\dfrac{128}{x^2}"

Find the critical number(s)


"A'(x)=0=>4-\\dfrac{128}{x^2}=0"

"x^2=32"

"x_1=-4\\sqrt{2}, x_2=4\\sqrt{2}"

If "0<x<4\\sqrt{2}, A'(x)>0, A(x)" increases.


If "x>4\\sqrt{2}, A'(x)<0, A(x)" decreases.


The function "A(x)" has a local maximum at "x=4\\sqrt{2}."

Since the function "A(x)" has the only extremum for "x>0," then the function "A(x)" has the absolute maximum at "x=4\\sqrt{2}" for "x>0."



"y=\\dfrac{32}{4\\sqrt{2}}=4\\sqrt{2}"

The poster is "(4\\sqrt{2}+4)\\ inches\\times (4\\sqrt{2}+4)\\ inches."


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