A rectangular storage container with an open top is to have a volume of 10 m3. The length of this base is twice the width. Material for the base costs $20 per square meter. Material for the sides costs $12 per square meter. Find the cost of materials for the cheapest such container. (Round your answer to the nearest cent.) $

Answer:$ 327.08Step-by-step explanation:Let w be the width ( in meters ) of the container,⇒ Length of the container = 2w,If h be the height of the container,So, the volume of the container = length × width × height= 2w × w × h= 2w² hAccording to the question,[tex]2w^2h=10[/tex][tex]\implies h=\frac{10}{2w^2}=\frac{5}{w^2}[/tex]Now, the area of the base = length × width [tex]=2w^2[/tex]Area of sides = 2 × length × height + 2 × width × height[tex]=2\times 2w\times h+2\times w\times h[/tex][tex]=4w\times \frac{5}{w^2}+2w\times \frac{5}{w^2}[/tex][tex]=\frac{20}{w}+\frac{10}{w}[/tex][tex]=\frac{30}{w}[/tex]Since, material for the base costs $20 per square meter and material for the sides costs $12 per square meter,Hence, total cost,[tex]C(w) = 2w^2\times 20 +\frac{30}{w}\times 12[/tex][tex]C(w)=40w^2+\frac{360}{w}[/tex] Differentiating with respect to w,[tex]C'(w) = 80w - \frac{360}{w^2}[/tex]Again differentiating with respect to w,[tex]C''(w) = 80 +\frac{720}{w^3}[/tex]For maxima or minima,C'(w) = 0[tex]80w - \frac{360}{w^2}=0[/tex][tex]80w^3-360=0[/tex][tex]80w^3=360[/tex][tex]\implies w=\sqrt[3]{\frac{360}{80}}=1.65096362445\approx 1.651[/tex]For w = 1.651, C''(w) = positive,Thus, cost is minimum for width 1.651 meters,And, the minimum cost = C(1.651) = [tex]40(1.651)^2+\frac{360}{1.651}=\$327.081706869\approx \$ 327.08[/tex]