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HW22 Optimization: Problem 1
(1 point)

If 1100 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box

Volume $=$
(include units)
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Step 1 ：Let the side length of the square base be $x$ and the height of the box be $h$.

Step 2 ：The volume of the box is $V=x^{2}h$.

Step 3 ：The surface area of the material is $A=x^2+4xh=1100$ square centimeters.

Step 4 ：Solve for $h$ in terms of $x$ using the surface area equation: $h=\frac{1100-x^2}{4x}$.

Step 5 ：Substitute $h$ into the volume equation to get $V(x)=x^2\left(\frac{1100-x^2}{4x}\right)$.

Step 6 ：Simplify the volume function to $V(x)=\frac{1100x-x^3}{4}$.

Step 7 ：Take the derivative of $V(x)$ with respect to $x$ and set it equal to zero to find the critical points.

Step 8 ：Solve $\frac{dV}{dx}=0$ to find the optimal value of $x$ that maximizes the volume.

Step 9 ：The optimal side length of the base is $x_{\text{optimal}}=19.148541073664777$ centimeters.

Step 10 ：The optimal height of the box is $h_{\text{optimal}}=9.574272159025412$ centimeters.

Step 11 ：The maximum volume of the box is $V_{\text{max}}=3510.566061773223$ cubic centimeters.

Step 12 ：The largest possible volume of the box with a square base and an open top, using 1100 square centimeters of material, is $\boxed{{\displaystyle 3510.566061773223}}$ cubic centimeters.