Question

1. -/1 pointsa rectangular storage container without a lid is to have a volume of 10 m3. the length of its base is twice the width. material for the base costs $15 per square meter. material for the sides costs $9 per square meter.find the cost (in dollars) of materials for the least expensive such container. (round your answer to the nearest cent.)$

Explanation:

Step1: Let the width be \(x\)

Let the width of the base be \(x\) meters. Then the length of the base is \(2x\) meters. Let the height be \(h\) meters. The volume \(V = l\times w\times h\), and \(V = 10\) \(m^{3}\). So \(10=(2x)\times x\times h\), and \(h=\frac{10}{2x^{2}}=\frac{5}{x^{2}}\).

Step2: Find the cost function

The area of the base \(A_{base}=2x\times x = 2x^{2}\). The cost of the base \(C_{base}=15\times2x^{2}=30x^{2}\). The area of the four sides: two sides have area \(2x\times h\) and two sides have area \(x\times h\). The total area of the sides \(A_{sides}=2(2x\times h)+2(x\times h)=6xh\). The cost of the sides \(C_{sides}=9\times6xh = 54xh\). Substitute \(h = \frac{5}{x^{2}}\) into the cost function: \(C(x)=30x^{2}+54x\times\frac{5}{x^{2}}=30x^{2}+\frac{270}{x}\).

Step3: Take the derivative

Differentiate \(C(x)\) with respect to \(x\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), \(C^\prime(x)=60x-\frac{270}{x^{2}}\).

Step4: Set the derivative equal to zero

Set \(C^\prime(x) = 0\), so \(60x-\frac{270}{x^{2}}=0\). Multiply through by \(x^{2}\) to get \(60x^{3}-270 = 0\). Then \(x^{3}=\frac{270}{60}=\frac{9}{2}\), and \(x=\sqrt[3]{\frac{9}{2}}\approx1.65\).

Step5: Find the second - derivative

Differentiate \(C^\prime(x)\) to get \(C^{\prime\prime}(x)=60+\frac{540}{x^{3}}\). When \(x=\sqrt[3]{\frac{9}{2}}\), \(C^{\prime\prime}(x)>0\), so \(C(x)\) has a minimum at this \(x\).

Step6: Calculate the minimum cost

Substitute \(x=\sqrt[3]{\frac{9}{2}}\) into \(C(x)\). \(h=\frac{5}{x^{2}}=\frac{5}{(\frac{9}{2})^{\frac{2}{3}}}\). \(C(x)=30x^{2}+\frac{270}{x}\). \(x^{3}=\frac{9}{2}\), \(x^{2}=\frac{9}{2x}\). \(C(x)=30\times\frac{9}{2x}+\frac{270}{x}=\frac{135}{x}+\frac{270}{x}=\frac{405}{x}\). Since \(x=\sqrt[3]{\frac{9}{2}}\), \(C(x)\approx30\times(1.65)^{2}+\frac{270}{1.65}\approx30\times2.7225 + 163.64\approx81.675+163.64\approx245.32\).

Answer:

\(245.32\)