Art dealer question

Question

An art dealer advises that, for the best display of a painting, the area of the frame should be (1/5) the area of the painting. Determine the best width of the frame for a painting 20 cm by 20 cm, if the width is to be the same on all sides. Round your answer to the nearest tenth. (6 marks)

Please answer in full detail

Answer 1

Here is a sketch:

The area of the picture is (20)(20) = 400 cm²

The area of the total frame/picture is (20 + 2x)(20 + 2x) = 400 + 80x + 4x²

So the area of just the frame is (400 + 80x + 4x²) - (400) = 4x² + 80x

We want the area of the frame to be 1/5 the area of the picture so it must be (1/5)(400) = 80. That means:

4x² + 80x = 80

4x² + 80x - 80 = 0

x² + 20x - 20 = 0

Using the quadratic formula:

For ax² + bx + c = 0 then

x = [-b ± √(b² - 4ac)]/(2a)

So, since a = 1, b = 20, c = -20 then:

x = [-20 ± √(20² - 4(1)(-20))]/(2•1)

x = (-20 ± √480)/2

x = (-20 ± 21.91)/2

x = 0.95 or x = -20.95

We can throw out the negative answer so the frame width (x) is 0.95 cm or, rounded to the nearest tenth: 1.0 cm

Answer 2

area of the painting is 20^2= 400 cm^2

area of the frame = 400/5 = 80 cm^2

area of both

= (20+2x)(20+2x) = 480

= 400 +80x + 4x^2 -480=0

4x^2 + 80x -80 =0

x^2 +20x -20 = 0

x= -20/2 +(1/2)sqr(400+80)

x =-10 +sqr120

x= about 0.95 cm= about 1 cm width of the frame

picture plus frame = about 22 by 22 = aout 484 cm^2

picture = 400 cm^2

frame area = about 84 cm^2 = about 1/5 of 400 = 80 cm^2

width of the frame on top, bottom, left and right = 0.95 cm

to nearest tenth the width = 1.0 centimeters