Question

A box is to have square base an open top and volume of 32 meter cube.find the dimension of the box that box that use the last amount of material.

Answer 1

Answer on Question #66744 – Math – Differential Equations

Question

A box is to have square base an open top and volume of 32 meter cube. Find the dimension of the box that uses the least amount of material.

Solution

$$V = x ^ {2} h = 32 \rightarrow h = \frac {32}{x ^ {2}}.$$ $$S = x ^ {2} + 4 x h = x ^ {2} + \frac {128}{x}.$$ $$\frac {dS}{dx} = 0 \rightarrow 2x - \frac {128}{x ^ {2}} = 0 \rightarrow x = \sqrt [3]{64} = 4.$$ $$\left. \frac {d ^ {2} S}{d x ^ {2}} (4) = \frac {256}{x ^ {3}} \right| _ {x = 4} = \frac {256}{4 ^ {3}} > 0.$$

So $S$ has minimum at $x_{*} = 4$ and

$$h _ {*} = \frac {32}{x _ {*} ^ {2}} = \frac {32}{4 ^ {2}} = \frac {32}{16} = 2.$$

The box uses the least amount of material when $x = 4 \, \text{m}$, $h = 2 \, \text{m}$.

Answer: $4 \, \text{m}, 4 \, \text{m}, 2 \, \text{m}$.

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