Respuesta :
Answer:
$ 327.08
Step-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 h
According 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]
The volume of a container is the amount of space in it.
The cost of materials for the cheapest such container is $327.08
The volume of the container is given as:
[tex]V = 10m^3[/tex]
The length is twice the width;
This means that:
[tex]l =2w[/tex]
The volume of the container is calculated as:
[tex]V = lwh[/tex]
Substitute [tex]l =2w[/tex]
[tex]V = 2w\times wh[/tex]
[tex]V = 2w^2h[/tex]
Substitute [tex]V = 10m^3[/tex]
[tex]2w^2h = 10[/tex]
Divide through by 2
[tex]w^2h = 5[/tex]
Make h the subject
[tex]h = \frac{5}{w^2}[/tex]
The surface area of an open rectangular storage is:
[tex]A =Area\ of\ sides + Area\ of\ base[/tex]
This gives
[tex]A = 2h(l + w) + lw[/tex]
Substitute [tex]l =2w[/tex]
[tex]A = 2h(2w + w) + 2w \times w[/tex]
[tex]A = 2h(2w + w) + 2w^2[/tex]
[tex]A =2h(3w) + 2w^2[/tex]
[tex]A =6hw + 2w^2[/tex]
Substitute [tex]h = \frac{5}{w^2}[/tex]
[tex]A =6w \times \frac{5}{w^2} + 2w^2[/tex]
[tex]A = \frac{30}{w} + 2w^2[/tex]
The base costs $20, and the sides cost $12
So, the cost function is:
[tex]C(w) =12 \times \frac{30}{w} + 2w^2 \times 20[/tex]
[tex]C(w) = \frac{360}{w} + 40w^2[/tex]
Rewrite as:
[tex]C(w) = 360w^{-1}+ 40w^2[/tex]
Differentiate
[tex]C'(w) = -360w^{-2}+ 80w[/tex]
Set to 0
[tex]-360w^{-2}+ 80w = 0[/tex]
Rewrite as:
[tex]-360w^{-2}=- 80w[/tex]
Cancel out the negatives
[tex]360w^{-2}=80w[/tex]
Multiply through by w^2
[tex]360=80w ^3[/tex]
Divide through by 80
[tex]4.5=w ^3[/tex]
Take cube roots of both sides
[tex]w = 1.65[/tex]
Recall that:
[tex]C(w) = 360w^{-1}+ 40w^2[/tex]
So, we have:
[tex]C(1.65) = 360 \times 1.65^{-1} + 40 \times 1.65^2[/tex]
[tex]C(1.65) = 327.08[/tex]
Hence, the cost of materials for the cheapest such container is $327.08
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