Hodgson Ellis

07/06/2023 · Middle School

\begin{tabular}{l} A box with a square base and open top must have a volume of \( 470596 \mathrm{~cm}^{3} \). 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. \\ \( A(x)=x^{2}+\frac{1882384}{x} \) \\ Next, find the derivative, \( A^{\prime}(x) \). \\ \( A^{\prime}(x)=2 x-\frac{1882384}{x^{2}} \) \\ Now, calculate when the derivative equals zero, that is, when \( A^{\prime}(x)=0 \). [Hint: multiply both sides by \( x^{2} \).] \\ \( A^{\prime}(x)=0 \) when \( x=98.15 \) \\ We next have to make sure that this value of \( x \) gives a minimum value for the surface area. Let's use the \\ second derivative test. Find \( A^{\prime \prime}(x) \). \\ \( A^{\prime \prime}(x)=2+\frac{2 \cdot 1882384}{x^{3}} \) \\ Evaluate \( A^{\prime \prime}(x) \) at the \( x \)-value you gave above. \\ \hline 98.15 \\ 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 \( A^{\prime}(x) \) must indicate a local minimum for \( A(x) \). (Your boss is happy now.) \end{tabular}