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 open - top box \(A(x)=x^{2}+4xh\). Substitute \(h=\frac{48668}{x^{2}}\) into the surface area formula:
\(A(x)=x^{2}+4x\times\frac{48668}{x^{2}}=x^{2}+\frac{194672}{x}\)

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

Using the power rule \((x^{n})^\prime=nx^{n - 1}\), \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\) (or rewrite \(\frac{194672}{x}=194672x^{-1}\)).
\(A^{\prime}(x)=2x-\frac{194672}{x^{2}}\)

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

Set \(2x-\frac{194672}{x^{2}}=0\). Multiply through by \(x^{2}\) (since \(x
eq0\), as \(x\) represents a length).
We get \(2x^{3}-194672 = 0\), then \(x^{3}=\frac{194672}{2}=97336\), so \(x=\sqrt[3]{97336}=46\)

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

Differentiate \(A^{\prime}(x)=2x - 194672x^{-2}\)
\(A^{\prime\prime}(x)=2+\frac{389344}{x^{3}}\)

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

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

Answer:

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