Precalculus question

Question

A graphic designer wants to create a rectangular graphic that has a 2 -inch margin on each side and a 4-inch margin on the top and the bottom. The design, including the margins, should have an area of 392 square inches. What overall dimensions will maximize the size of the design , excluding the margins?

Answer 1

Draw the figure: One large rectangle with a smaller one inside.

Label the dimensions of the inside rectangle x and y.

Then the dimensions of the larger rectangle are x + 4 and y + 8.

It is given that the area of the large rectangle should be 392. So....

(x + 4) ( y + 8) = 392

We want to maximize the area of the interior rectangle. The formula will be

A = x y

So we begin by expressing A in terms of x. So solve the other equation for y.

(x + 4) ( y + 8) = 392. Divide both sides by (x + 4)

y + 8 = 392 / (x + 4) Subtract 8 from both sides.

y = [392 / (x + 4)] - 8 Put both terms over the common denominator of (x + 4)

y = 392 - 8 (x + 4) Simplify the numerator.

(x + 4)

y = 392 - 8x - 32

x + 4

y = -8x + 360

x + 4

Substituting this value into the Area equation.

A = x (-8x + 360) or. -8x² + 360x This is what we must maximize. So take its first derivative.

x + 4 x + 4 This uses the quotient rule.

A' = (x + 4) (-16x + 360) - (( -8x² + 360x) (1)

(x + 4) ²

A' = -16x² + 360x - 64x + 1440 + 8x² - 360x Multiplying out the numerator, not the denominator.

(x + 4)²

A' = -8x² - 64x + 1440 Collecting like terms.

(x + 4)² Set the numerator = 0 to find max and min.

A' = -8x² - 64x + 1440 = 0 Divide by -8

x² + 8x - 180 = 0 Factor

(x + 18)(x - 10) = 0 Solve for x. Discard the negative value.

x = -18 x = 10 Substitute that value into above equation for y.

y = -8x + 360 = -8(10)+360 = 280 = 20

x + 4 10 + 4 14

So the dimensions of the INTERIOR design is 10 by 20 and the dimensions of the overall dimensions will be

14 by 28. Check: Is 14 X 28 = 392???? YES!!!