college algebra

Question

A rectangular piece of metal is 15 in longer than it is wide. Squares with in long are cut from the four corners and the flaps are folded upward to form an open box. If the volume of the box is 1350 in3​, what were the original dimensions of the piece of​ metal?

Answer 1

It looks like you're missing the length of the sides of the squares that are cut from your rectangle. I'll show you how to solve it in general and you can just plug in the length of the square according to the problem you're working with to get your answer.

Start out by drawing a rectangle. Actually draw this picture it will really help. Label the width of the rectangle w, and the length w+15.

Now draw a square in each of the four corners of the rectangle. All of the squares will have the same side length, I will call it x.

When we remove the squares we now have a sort of cross shape. The new length of this shape is w+15-2x (since we remove two squares from the length each of length x). The new width of this shape is w-2x. The places where we cut into the rectangle all have length of x because we removed squares with side length x.

If you think about the rectangle that forms the base it will have length w+15-2x and width w-2x. The overall box will have a height of x. Therefore the volume formula is V = (w+15-2x)(w-2x)(x)

So if they tell you that the volume is 1350 just plug that in where V is in the equation. Your goal is to solve for w (the width of the original rectangle). Once you do that you can find the length by plugging w into w+15. In order to do this though you need to know the value of x (the side length of the squares that were removed).

Plug in given x (square side length) and solve for w (rectangle width): 1350 = (w+15-2x)(w-2x)(x)

You should end up solving a quadratic equation for w. If you end up with w = - # reject that root because we cannot have a negative length. Only take positive values for dimensions as possible answers.

Hope this helps!