Question
Jan Villaroel
Topic: Calculus Posted 1 year ago
An open rectangular box with a square base is to be made from 36 square meters of material. Find the dimensions that minimize the surface area.

An open rectangular box with a square base is to be made from 36 square meters of material. Find the dimensions that minimize the surface area.


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Koala Tutor
Nov 1, 2023
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Let's denote the side length of the square base as $x$, and the height of the rectangular box as $h$.

We know that the surface area of the box is given by $S = 2x^2 + 4xh$. We are also given that the material used to make the box has an area of 36 square meters, so we have the constraint $x^2 + 2xh = 36$.

To minimize the surface area, we need to find the critical points of the surface area function $S(x, h)$ subject to the constraint $x^2 + 2xh = 36$.

From the constraint, we can solve for $h$ in terms of $x$ as $h = \frac{36 - x^2}{2x}$ and substitute this into the surface area function: $S(x) = 2x^2 + 4x\left(\frac{36 - x^2}{2x}\right)$.

Simplifying this expression gives us $S(x) = 2x^2 + 2(36 - x^2) = 4x^2 + 72$.

To find the critical points, we take the derivative of $S(x)$ with respect to $x$ and set it to zero: $S'(x) = 8x = 0$. This gives us $x = 0$, but this is not a valid solution since we are looking for the dimensions of a physical box.

Thus, there are no critical points in the interior of the feasible region.

Next, we check the boundary of the feasible region. The constraint equation gives us $h = \frac{36 - x^2}{2x}$.

Taking the surface area function and substituting this value of $h$ into it gives us $S(x) = 2x^2 + 4x\left(\frac{36 - x^2}{2x}\right) = 2x^2 + 4(36 - x^2) = -2x^2 + 144$.

Now we need to find the critical points of $S(x)$ restricted to the boundary of the feasible region. Since the inequality sign does not change, the feasible region is compact because it's the intersection of a closed plane and a closed half-space.

Let's find the endpoints of this bounded region. Setting $x^2 + 2xh = 36$ to zero gives us $x^2 + 2xh - 36 = 0$, which factors to $(x+6)(x-6) = 0$.

So, $x = -6$ is not meaningful, but $x = 6$ gives us a base square with side length 6 meters.

The endpoints of the feasible region are $x = 6$ and $h = \frac{36 - x^2}{2x} = 3$.

Calculating the surface areas at these endpoints, we get $S(6) = -2(6)^2 + 144 = 72$.

Therefore, the dimensions that minimize the surface area are a base square with side length 6 meters and a height of 3 meters.

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