algebra question

Question

You are designing a rectangular swimming pool that is to be set into the ground. The width of the

pool is 5 feet more than the depth, and the length is 35 more feet than the depth. The pool holds 2000

cubic feet of water. What are the dimensions of the pool?

Answer 1

We are given the total volume of the pool which is 2000 cubic feet of water. We can begin by creating an equation that shows the relationship of the volume of the pool.

Since the pool itself is rectangular, we know the volume to be: V = LxWxH which in this case, H (height) is Depth. So we can rewrite the equation as: V = LxWxD.

We are told that the width is 5 feet more than the depth or: w = 5 + d.

We are told that the length is 35 feet more than the depth or: L = 35 + d

We can safely assume depth will simply just be denoted as the variable "d".

Knowing this, we can substitute in the variables w and L into our volume equation: V = (35 + d)(5+d)(d). We also know the total volume so we can rewrite once more as: 2000 = (35 + d)(5+d)(d).

Now we can begin by foiling the right hand side of the equation to get: 2000 = 175d + 400d² + d³. To solve for this, we will need to move the 2000 to the other side to set the equation equal to zero: 0 = d³ + 40d² + 175d -2000. By using the rational zeroes theorem, we can find that the only positive real solution to this third order polynomial is d = 5.

Now that we know that the depth is 5, we can find the dimensions of the pool which is LxWxD. So we get: (35+5)ft x (5+5)ft x 5ft which finally becomes: 40ft x 10ft x 5ft. This is the final dimensions of the pool that will give us the total volume of 2000 ft³.