Question

NO LINKS!!

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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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

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

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.