Answer to Question #122946 in Calculus for Kylie

Question

A friend asked you for some help in designing a poster. The poster your friend is designing must contain 50 square inches of printing with margins of 4 inches at the top and bottom, and 2 inch margins on each side. Your friend wants to use the least amount of paper possible. What overall dimensions (length and width) should you tell your friend to design the poster with in order to make this possible?
Draw a sketch of the situation and label all given quantities.

Answer 1

Let length and breadth of printing area are y inches and x inches respectively.

Given that printing area= 50 square inches

So xy = 50

Now length of paper = y+4+4 = y+8 inches

Breadth of paper = x+2+2 = x+4 inches

Area of paper , A = (x+4)(y+8)

= xy + 8x + 4y + 32

= 50 + 8x + 4y + 32 Since xy = 50

= 8x + 4y + 82

= 8x + 4* $\frac {50}{x}$ + 82 Since y = $\frac {50}{x}$

= 8x + $\frac {200}{x}$ + 82

Now we are to minimise A

$\frac {dA}{dx}$ = 8 - $\frac {200}{x^2}$

$\frac {d²y}{dx²}$ = $\frac {400}{x^3}$

For extreme values of A, $\frac {dA}{dx}$ = 0

=> 8 - $\frac {200}{x^2}$ = 0

=> 8x² = 200

=> x² = 25

=> x = 5 as x can not be negative

$[\frac {d²y}{dx²}]$_{x = 5} $=\frac {400}{5³} = \frac {400}{125} = \frac {16}{5} >0$

So A is minimum when x= 5

When x= 5, y = $\frac {50}{5}$ = 10

Therefore dimensions of printing are to use least amount of paper as follows.

Length of printing area = 10 inches

Breadth of printing area = 5 inches

Consequently overall dimensions of paper to use least amount, are as follows

Length of overall paper = y+8 = 10+8 = 18 inches

Breadth of overall paper = x+4 = 5+4 = 9 inches