A waste management company is designing a rectangular construction dumpster that will be twice as long as it is wide and must hold $10 \mathrm{yd}^{3}$ of debris. Find the dimensions of the dumpster that will minimize its surface area.
Write the surface area formula in terms of the width, $x$. Assume the dumpster has an open top.
\[
\mathrm{SA}=
\]

Step 1 ：The dimensions of the dumpster are twice as long as it is wide, so the length is \(2x\), and the width is \(x\). Let's denote the height as \(h\).

Step 2 ：The volume of the dumpster is given by \(V = lwh = 2x \cdot x \cdot h = 2x^2h\). We know that the volume must be \(10 \, yd^3\), so we can solve for \(h\) to get \(h = \frac{10}{2x^2} = \frac{5}{x^2}\).

Step 3 ：The surface area of the dumpster (with an open top) is given by \(SA = lw + 2lh + 2wh = 2x^2 + 4xh + 2xh = 2x^2 + 6xh\). Substituting \(h = \frac{5}{x^2}\) into the surface area formula, we get \(SA = 2x^2 + 6x \cdot \frac{5}{x^2} = 2x^2 + \frac{30}{x}\).

Step 4 ：To minimize the surface area, we take the derivative of the surface area with respect to \(x\) and set it equal to zero. This gives us \(\frac{dSA}{dx} = 4x - \frac{30}{x^2} = 0\).

Step 5 ：Solving for \(x\), we get \(x^3 = \frac{30}{4} = 7.5\), so \(x = \sqrt[3]{7.5} \approx 1.957\).

Step 6 ：We check the second derivative to confirm that this is a minimum. The second derivative is \(\frac{d^2SA}{dx^2} = 4 + \frac{60}{x^3}\), which is positive for \(x > 0\), so \(x = \sqrt[3]{7.5}\) indeed gives a minimum surface area.

Step 7 ：Substituting \(x = \sqrt[3]{7.5}\) back into the formula for \(h\), we get \(h = \frac{5}{x^2} = \frac{5}{(\sqrt[3]{7.5})^2} \approx 1.3\).

Step 8 ：So the dimensions that will minimize the surface area of the dumpster are length \(2x = 2\sqrt[3]{7.5} \approx 3.914 \, yd\), width \(x = \sqrt[3]{7.5} \approx 1.957 \, yd\), and height \(h = \frac{5}{x^2} \approx 1.3 \, yd\).