Answer:
The least expensive cost = $245.32
Step-by-step explanation:
From the information:
Volume of the rectangular container = 10 m²
Assume the width of the container is = w
The base = 2 × width
= 2(w)
Volume = height × base × width
10 = h × 2w × w
[tex]h = \dfrac{10}{2w^2}[/tex]
[tex]h = \dfrac{5}{w^2}[/tex]
Now, to calculate the area of the base; we have:
[tex]Area = length \times width[/tex]
[tex]Area = 2w \times w[/tex]
[tex]Area = (2w^2) \ m^2[/tex]
[tex]Cost \ of \ base = 15 * 2w^2[/tex]
Cost of base = [tex]30w^2[/tex] --- (1)
[tex]Area \ of \ side = 2 ( height \times length ) + 2( height \times width) \\ \\ = 2 (h \times 2w) + 2(h\times w) \\ \\ = 4hw + 2hw \\ \\ = 6 hw \\ \\ = 6 \times \dfrac{5}{w^2} \times w \\ \\ Area = \dfrac{30 }{w} \ m^2[/tex]
Let's not forget that the cost of the sides = $9 per m²
Thus; the cost of the sides will be [tex]= \$ 9 \times \dfrac{30}{w}[/tex]
the cost of the sides [tex]= \dfrac{\$270}{w}[/tex]
Thus the total cost = cost of base = cost of sides
[tex]C(w) = 30w^2 + \dfrac{270}{w}[/tex]
By differentiation;
[tex]C'(w) = \dfrac{dC}{dw} \\ \\ = 30 \times 2w + 270 ( \dfrac{-1}{w^2}) \\ \\ C'(w) = 60 w - \dfrac{270 }{w^2} \\ \\[/tex]
[tex]\text{To determine the minimum cost:} \\ \\ C'(w) = 0 \\ \\ So, 60 w - \dfrac{270}{w^2}= 0 \\ \\ \text{By integrating w.r.t w} \\ \\ w ( 60 - \dfrac{270}{w^3}) =0 \\ \\ \implies 60 - \dfrac{270}{w^3} \\ \\ w^3 = \dfrac{270}{60} \\ \\ w^3 = \dfrac{9}{2} \\ \\ w = (\dfrac{9}{2})^{1/3} \\ \\ w = 1.6509 \ m[/tex]
From; [tex]C(w) = 30w^2 + \dfrac{270}{w}[/tex]
[tex]Cos t = \dfrac{30 w^3+ 270}{w} \\ \\ =\dfrac{30 (\dfrac{9}{2})+ 270}{1.6509} \\ \\ = \dfrac{135+270}{1.6509} \\ \\ = \dfrac{405}{1.6509} \\ \\ \mathbf{Cost = \$245.32}[/tex]
