Question

An artist wanted to paint a picture on a canvas which would allow for a margin of 4 inches on top and bottom and 2 inches on each side.
He wanted the picture itself to occupy 72 sq. inches.
What would be the smallest dimensions, the canvas he is going to obtain, should possess ?
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Answer 1

Answer:

A= 72 sq. inches.

x is the total length of the sides

One side will be equal to s1 = x-2-2= x-4 (inches)

Other side will be equal to s2= x-4-4=x-8 inches

A= s1*s2

72= (x-4)(x-8)

72= x^2-4x-8x+32

x^2-4x-8x+32-72=0

x^2-12x-40=0

x^2-12x=40

By completing square

Adding 6^2 to both sides

x^2-2(x)(6) +6^2= 40 + 6^2

(x-6)^2= 40+36

(x-6)^2=76

x-6= (76)^1/2

x= 4(19)^1/2+6

x= 14.72 inches

s1= x-4 = 14.72-4= 10.72 inches

s2= x-8 = 14.72-8= 6.72 inches

Let's check it

A= s1*s2= 10.72*6.72= 72 inches^2