Question

1. minimum - surface - area box of all boxes with a square base and a volume of 8m³, which one has the minimum surface area? (give its dimensions.)

Explanation:

Step1: Define variables

Let the side length of the square base be \(x\) (in meters) and the height of the box be \(h\) (in meters). The volume \(V=x^{2}h\), and since \(V = 8\), we have \(h=\frac{8}{x^{2}}\). The surface - area formula \(S=2x^{2}+4xh\).

Step2: Substitute \(h\) into the surface - area formula

Substitute \(h=\frac{8}{x^{2}}\) into \(S\): \(S(x)=2x^{2}+4x\cdot\frac{8}{x^{2}}=2x^{2}+\frac{32}{x}\), where \(x>0\).

Step3: Find the derivative of \(S(x)\)

Using the power rule, \(S^\prime(x)=4x-\frac{32}{x^{2}}\).

Step4: Set the derivative equal to zero and solve for \(x\)

Set \(S^\prime(x) = 0\):

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Step5: Find the second - derivative of \(S(x)\)

\(S^{\prime\prime}(x)=4+\frac{64}{x^{3}}\). When \(x = 2\), \(S^{\prime\prime}(2)=4+\frac{64}{8}=4 + 8=12>0\). So \(S(x)\) has a minimum at \(x = 2\).

Step6: Find the value of \(h\)

When \(x = 2\), \(h=\frac{8}{2^{2}}=2\).

Answer:

The box with dimensions \(2\space m\times2\space m\times2\space m\) (a cube) has the minimum surface area.