Question

a box with a square base and open top must have a volume of 48668 cm³. we wish to find the dimensions of the box that minimize the amount of material used. first, find a formula for the surface area of the box in terms of only x, the length of one side of the square base. hint: use the volume formula to express the height of the box in terms of x. simplify your formula as much as possible. a(x) = next, find the derivative, a(x). a(x) = now, calculate when the derivative equals zero, that is, when a(x) = 0. hint: multiply both sides by x². a(x) = 0 when x = we next have to make sure that this value of x gives a minimum value for the surface area. lets use the second derivative test. find a(x). a(x) = evaluate a(x) at the x - value you gave above. note: since your last answer is positive, this means that the graph of a(x) is concave up around that value, so the zero of a(x) must indicate a local minimum for a(x). (your boss is happy now.) question help: message instructor

Explanation:

Step1: Find the height in terms of \(x\)

The volume of the box \(V=x^{2}h\), given \(V = 48668\), so \(h=\frac{48668}{x^{2}}\).
The surface area of the box (open - top) \(A(x)=x^{2}+4xh\). Substitute \(h=\frac{48668}{x^{2}}\) into the surface - area formula:

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Step2: Find the first derivative \(A^{\prime}(x)\)

Using the power rule \((x^{n})^\prime=nx^{n - 1}\), for \(y = x^{2}+\frac{194672}{x}=x^{2}+194672x^{-1}\)
\(A^{\prime}(x)=2x-194672x^{-2}=2x-\frac{194672}{x^{2}}\)

Step3: Find when \(A^{\prime}(x) = 0\)

Set \(A^{\prime}(x)=0\), then \(2x-\frac{194672}{x^{2}} = 0\). Multiply both sides by \(x^{2}\) (since \(x\gt0\) as it represents a length)
\(2x^{3}-194672 = 0\), \(x^{3}=\frac{194672}{2}=97336\), \(x=\sqrt[3]{97336}=46\)

Step4: Find the second derivative \(A^{\prime\prime}(x)\)

Differentiate \(A^{\prime}(x)=2x - 194672x^{-2}\) with respect to \(x\). Using the power rule, \(A^{\prime\prime}(x)=2 + 2\times194672x^{-3}=2+\frac{389344}{x^{3}}\)

Step5: Evaluate \(A^{\prime\prime}(x)\) at \(x = 46\)

Substitute \(x = 46\) into \(A^{\prime\prime}(x)\): \(A^{\prime\prime}(46)=2+\frac{389344}{46^{3}}\)
\(46^{3}=46\times46\times46 = 97336\), \(A^{\prime\prime}(46)=2+\frac{389344}{97336}=2 + 4=6\)

Answer:

\(A(x)=x^{2}+\frac{194672}{x}\)
\(A^{\prime}(x)=2x-\frac{194672}{x^{2}}\)
\(A^{\prime}(x)=0\) when \(x = 46\)
\(A^{\prime\prime}(x)=2+\frac{389344}{x^{3}}\)
\(A^{\prime\prime}(46)=6\)