box costs $60 \mathrm{c} / \mathrm{ft}^{2}$. The material for the sides of the box costs $40 \mathrm{c} / \mathrm{ft}^{2}$. Find the dimensions of the least expensive box that can be constructed.
length
$\mathrm{ft}$
width
ft
height
ft
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Answer
Final Answer: The dimensions of the least expensive box that can be constructed are \(\boxed{2^{1/3} \cdot 3^{2/3} \cdot V^{1/3}/3}\) feet for the length, width, and height.
Step 1 ：The problem is asking for the dimensions of the least expensive box that can be constructed given the cost of the materials for the top/bottom and sides of the box. This is an optimization problem, which can be solved using calculus.
Step 2 ：First, we need to express the cost of the box as a function of its dimensions. The cost of the box is the sum of the cost of the top/bottom and the cost of the sides. The area of the top/bottom is length*width and the area of the sides is 2*(length*height + width*height).
Step 3 ：Next, we need to express one of the dimensions in terms of the other two. We can do this by using the volume of the box, which is length*width*height. Let's assume that the volume of the box is a constant value V. Then we can express height as V/(length*width).
Step 4 ：Finally, we can substitute the expression for height into the cost function and differentiate it with respect to one of the other dimensions. Setting the derivative equal to zero will give us the dimensions that minimize the cost.
Step 5 ：The solution to the system of equations gives three pairs of values for length and width. However, the second and third pairs are complex numbers, which do not make sense in the context of this problem. Therefore, the only valid solution is the first pair, which gives the same value for length and width. This means that the least expensive box that can be constructed is a cube.
Step 6 ：To find the height of the box, we can substitute the values of length and width into the expression for height.
Step 7 ：The height of the box is also equal to the length and width. This confirms that the least expensive box that can be constructed is a cube.
Step 8 ：Final Answer: The dimensions of the least expensive box that can be constructed are \(\boxed{2^{1/3} \cdot 3^{2/3} \cdot V^{1/3}/3}\) feet for the length, width, and height.
