A rectangular prism has a volume of 48 cubic feet and surface area of

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To find the dimensions of a rectangular prism given its volume and surface area, we need to use a system of equations.

Let's assume the length, width, and height of the rectangular prism are represented by L, W, and H, respectively.

The volume of a rectangular prism is given by the formula V = L * W * H. In this case, the volume is given as 48 cubic feet:

48 = L * W * H ----(Equation 1)

The surface area of a rectangular prism is given by the formula SA = 2(LW + LH + WH). In this case, the surface area is given as 88 square feet:

88 = 2(LW + LH + WH) ----(Equation 2)

We have two equations (Equation 1 and Equation 2) and three unknowns (L, W, and H). To solve for the dimensions, we can use the method of substitution.

Rearrange Equation 1 to solve for H:

H = 48 / (L * W) ----(Equation 3)

Substitute Equation 3 into Equation 2:

88 = 2(LW + L(48 / (L * W)) + W(48 / (L * W)))

Simplify:

88 = 2(LW + 48 / W + 48 / L)

Divide both sides of the equation by 2:

44 = LW + 24 / W + 24 / L ----(Equation 4)

Now we have a single equation with two variables. We can solve for either L or W and find the other dimension using the values obtained.

Let's solve Equation 4 for L:

44 = LW + 24 / W + 24 / L

Rearrange the equation:

LW = 44 - 24 / W - 24 / L

Multiply both sides by L:

L^2W = 44L - 24 - 24(W / L)

Rearrange the equation:

L^2W + 24(W / L) + 24 = 44L ----(Equation 5)

Now we have an equation that only involves L and W. We can use numerical methods like graphing, substitution, or trial and error to solve it.

Alternatively, you can use online software or a calculator capable of solving equations to find the possible values of L and W. By inputting Equation 5 into such tools, you can obtain the dimensions of the rectangular prism.

There are many solutions possible. However, assuming you are doing factoring in math, we will limit ourselves with integral sides, i.e. the length of each side is an integer.

If we factor 48 into its factors, we get:
2,2,2,2,3
The possible sides formed by these factors, ignoring sides of unity (1), are:
2,3,6,8,16,24,...
By trial and error, we start with
2,3,8: V=48, A=2(2*3+3*8+8*2)=92 too big.
We try sides that are more equal to each other:
2,4,6: V=48, A=2(2*4+4*6+6*2)=88 OK
So one possible solution for the sides is 2,4 and 6.