Question

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The length of a rectangle is 5 less than twice the width. What values of the width will make the area less than 150 square feet? Define your variable(s). Write inequalities that could be used to solve this problem. Then solve the problem.

Answer 1

Answer: Anything between 0 and 10, excluding both endpoints.

In terms of symbols we can say 0 < w < 10 where w is the width.

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Step-by-step explanation:

You could do this with two variables, but I think it's easier to instead use one variable only. This is because the length is dependent on what you pick for the width.

w = width

2w = twice the width

2w-5 = five less than twice the width = length

So,

• width = w
• length = 2w-5

which lead to

area = length*width

area = (2w-5)*w

area = 2w^2-5w

area < 150

2w^2 - 5w < 150

2w^2 - 5w - 150 < 0

To solve this inequality, we will solve the equation 2w^2-5w-150 = 0

Use the quadratic formula. Plug in a = 2, b = -5, c = -150

`w = (-b\pm√(b^2-4ac))/(2a)\\\\w = (-(-5)\pm√((-5)^2-4(2)(-150)))/(2(2))\\\\w = (5\pm√(1225))/(4)\\\\w = (5\pm35)/(4)\\\\w = (5+35)/(4) \ \text{ or } \ w = (5-35)/(4)\\\\w = (40)/(4) \ \text{ or } \ w = (-30)/(4)\\\\w = 10 \ \text{ or } \ w = -7.5\\\\`

Ignore the negative solution as it makes no sense to have a negative width.

The only practical root is w = 10.

If w = 10 feet, then the area = 2w^2-5w results in 150 square feet.

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Based on that root, we need to try a sample value that is to the left of it.

Let's say we try w = 5.

2w^2 - 5w < 150

2*5^2 - 5*5 < 150

25 < 150 ... which is true

This shows that if 0 < w < 10, then 2w^2-5w < 150 is true.

Now try something to the right of 10. I'll pick w = 15

2w^2 - 5w < 150

2*15^2 - 5*15 < 150

375 < 150 ... which is false

It means w > 10 leads to 2w^2-5w < 150 being false.

Therefore w > 10 isn't allowed if we want 2w^2-5w < 150 to be true.