A closed box is to be rectangular solid with a square base. If the volume is 32in^3, determine the dimensions for which the surface area is minimum.
make a sketch
let the base be x by x, and the height be y
so x^2 y = 32
y = 32/x^2
SA = 2x^2 + 4xy
= 2x^2 + 4x(32/x^2)
= 2x^2 + 128/x
d(SA)/dx = 4x - 128/x^2 = 0 for a min of SA
4x = 128/x^2
4x^3 = 128
x^3 = 32
x = 32^(1/3) or appr 3.175
y = 32/3.175^2
Well , what do you know, it happens to be a perfect cube.
60.48
To determine the dimensions for which the surface area of the closed box is minimum, we need to find a relationship between the dimensions and the surface area.
Let's assume the length, width, and height of the box are represented by L, W, and H, respectively. Since the box has a square base, we can say L = W.
The volume of a rectangular solid is given by the formula V = L * W * H. Since the volume of the box is given as 32in^3, we have:
32 = L * L * H
32 = L^2 * H
Now, we need to find a relationship for the surface area. The surface area of a closed box can be calculated using the formula:
SA = 2(LW + LH + WH)
Substituting L = W into this formula, we get:
SA = 2(L^2 + LH + LH)
SA = 2L^2 + 4LH
We can simplify the surface area equation by substituting the value of H from the volume equation:
SA = 2L^2 + 4(32 / L^2)
SA = 2L^2 + 128 / L
To find the dimensions for which the surface area is minimum, we need to find the critical points of the surface area equation. We can do this by taking the derivative of SA with respect to L and setting it equal to zero:
dSA/dL = 4L - 128 / L^2 = 0
To solve for L, we'll multiply both sides by L^2 and rearrange the equation:
4L^3 - 128 = 0
4L^3 = 128
L^3 = 32
L = ∛32
L ≈ 3.174
Since L = W, the length and width of the box should be approximately 3.174 inches.
Now, we can find the value of H by substituting the value of L into the volume equation:
32 = (3.174)^2 * H
32 = 10.08H
H ≈ 3.174 inches
Therefore, the dimensions for which the surface area is minimum are approximately:
Length = Width = 3.174 inches
Height = 3.174 inches