Question

sarah wants to put three paintings on her living room wall. the length of the wall is 15 feet longer than its width. the length and width of the paintings are 3 feet and 4 feet, respectively.

which of the following inequalities can be used to solve for x, if sarah wants the combined area of the wall and the paintings to be at the most 202 feet square feet?

Answer 1

Let
x-----------> the width of the wall
x+15------> the length of the wall 

we know that
x*(x+15)+3*[3*4] <=202 ft²
x²+15x+36 <=202
x²+15x+36 -202<=0
x²+15x-166<=0

the answer is 
x²+15x-166<=0

using a graph tool
see the attached figure
The maximum value of x is x=7.408 ft

Answer 2

The inequality used to determine x is x² + 15x + 12 ≤ 202

Inequality

Inequality is an expression used to show the non equal comparison of two or more numbers and variables.

Let x represent the width of the wall. Hence

Length of wall = x + 15

Area of wall = x(x + 15) = x² + 15x

area of painting = 3 feet * 4 feet = 12 feet

Combined area of wall and painting = x² + 15x + 12

x² + 15x + 12 ≤ 202

The inequality used to determine x is x² + 15x + 12 ≤ 202

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