a waste management company is designing a rectangular construction dumpster that will be twice as long as it…

a waste management company is designing a rectangular construction dumpster that will be twice as long as it is wide and must hold 25 yd³ 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. sa = 2x² + 75/x the width of the dumpster is the length of the dumpster is the height of the dumpster is (round to the nearest hundredth as needed )

a waste management company is designing a rectangular construction dumpster that will be twice as long as it is wide and must hold 25 yd³ 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. sa = 2x² + 75/x the width of the dumpster is the length of the dumpster is the height of the dumpster is (round to the nearest hundredth as needed )

Answer

Explanation:

Step1: Find the derivative of the surface area function

Given (SA = 2x^{2}+\frac{75}{x}), the derivative (SA^\prime) using the power rule ((x^n)^\prime=nx^{n - 1}) is: (SA^\prime=4x-\frac{75}{x^{2}})

Step2: Set the derivative equal to zero and solve for (x)

Set (4x-\frac{75}{x^{2}} = 0). Multiply through by (x^{2}) to get (4x^{3}-75 = 0). Then (x^{3}=\frac{75}{4}), so (x=\sqrt[3]{\frac{75}{4}}\approx2.67)

Step3: Find the length

Since the length (l = 2x), substituting (x\approx2.67), we get (l\approx2\times2.67 = 5.34)

Step4: Find the height

The volume (V=l\times w\times h), and (V = 25), (l = 2x), (w=x). So (25=(2x)\times x\times h), (h=\frac{25}{2x^{2}}). Substituting (x\approx2.67), (h=\frac{25}{2\times(2.67)^{2}}\approx1.78)

Answer:

The width of the dumpster is (2.67) yd. The length of the dumpster is (5.34) yd. The height of the dumpster is (1.78) yd.