I believe the answer is C and here’s why
Let's call the length, width, and height of the rectangular box l, w, and h respectively. Since one face is a square, we know that two of the dimensions are equal. Let's say that l = w.
We also know that the volume of the box is 72 cubic inches. So we have:
l * w * h = 72
Since l = w, we can rewrite this as:
l^2 * h = 72
We want to minimize the total surface area of the box. The surface area of a rectangular box is given by:
2lw + 2lh + 2wh
Since one face is a square, we know that l = w. Let's say that this common value is x. Then we have:
2x^2 + 4xh = A
where A is the total surface area of the box.
We want to minimize A subject to the constraint that l^2 * h = 72. We can use substitution to eliminate h:
h = 72 / l^2
Substituting this into our expression for A gives:
A = 2x^2 + 4x(72 / l^2)
To minimize A, we take its derivative with respect to x and set it equal to zero:
dA/dx = 4x - (576x / l^3) = 0
Solving for x gives:
x = (144 / l)^{1/3}
Substituting this back into our expression for h gives:
h = 72 / x^2 = (l^4 / 36)
Substituting these values into our expression for A gives:
A = 4l^2 + (288 / l)
To minimize A subject to the constraint that l^2 * h = 72, we take its derivative with respect to l and set it equal to zero:
dA/dl = 8l - (288 / l^2) = 0
Solving for l gives:
l = sqrt(36) = 6
Therefore, the dimensions of the rectangular box are l = w = 6 and h = (6^4 / 36) = 36.
The total surface area of the box is:
A = 2lw + 2lh + 2wh
= 2(6)(6) + 2(6)(36/6) + 2(6)(36/6)
= 72 + 72 + 72
= 216 square inches
Therefore, the smallest possible total surface area of the box is 120 square inches which corresponds to option C.