Final answer:
To minimize the surface area of the box, the dimensions that will give the minimum surface area while maintaining a volume of 4 cubic inches are x = 2 inches and h = 1 inch. The minimum surface area is 16 square inches.
Step-by-step explanation:
To minimize the surface area of the open-top, square-based, rectangular box, we need to find the dimensions that give the minimum surface area while still maintaining a volume of 4 cubic inches. Let's denote the length of one side of the square base as x and the height of the box as h. The volume of the box is given by V = x^2 * h, and we want it to be equal to 4. Therefore, x^2 * h = 4.
Now, the surface area of the box can be calculated as follows: the top and bottom of the box each have an area of x^2, and the four sides each have an area of x * h. So the total surface area, A, is given by A = 2x^2 + 4xh.
To find the dimensions that minimize the surface area, we can use the volume equation to solve for h in terms of x: h = 4/x^2. Substituting this expression into the surface area equation, we have A = 2x^2 + 4x(4/x^2) = 2x^2 + 16/x.
Now, we can find the minimum surface area by taking the derivative of A with respect to x and setting it equal to zero:
dA/dx = 4x - 16/x^2 = 0.
Solving this equation, we find x^3 = 4, which implies x = 2. Substituting this value back into the volume equation, we can find the corresponding value of h: h = 4/2^2 = 1.
Therefore, the dimensions that minimize the surface area are x = 2 inches and h = 1 inch. Plugging these values into the surface area equation, we get A = 2(2^2) + 16/2 = 8 + 8 = 16 square inches.