McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
5. Quadratic Inequalities
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Exercise 32 Page 211

Express the volume as a perfect square.

8

Practice makes perfect

We want to find the dimensions of the box before the corners were cut.

We are given the formula for the volume of the box. V(x)=4x^2-64x+256Notice that every coefficient is divisible by 4, so let's factor it out. V(x)=4(x^2-16x+64) Now we can see that the quadratic is a perfect square. V(x)=4(x-8)^2 We are asked to find the values of x for which this volume does not exceed 750 cubic inches. Let's write and solve an inequality.
4(x-8)^2≤ 750
(x-8)^2≤ 187.5
|x-8|≤ sqrt(187.5)
We know that the meaningful values of x are greater than 8, because the box must be longer than 8cm if that much is cut from the corners. This means that x-8 is positive, so we do not need the absolute value. 0

Extra

Why is the volume formula given in the question?

Note that the diagram would have been enough to find the volume, and there was no need to give the formula in the question. Volume is the product of the dimensions, and the box has a (x-8)*(x-8) square base and is 4 inches high. V(x)=4(x-8)(x-8)=4(x-8)^2 The given expanded form was actually hiding this more useful form of the volume.