Answer :
Dimensions are 3.52m × 1.76m × 1.76m
minimum cost is $59.56
Let w be the width of the container.
Since, the length is twice of the width, so length of the container will be 2w.
Now, let h be the height of the container.
Volume = length × width × height
16 = 2w × w × h
8 = [tex]w^{2}[/tex] × h
h = [tex]\frac{8}{w^{2} }[/tex]
Since, the area of the base = 1 × w = 2w × w = [tex]2w^{2}[/tex]
Area of the lid = 1 × w = [tex]2w^{2}[/tex]
while the area of the sides = 2hw + 2hl
= 2h(w + l)
= 2 × [tex]\frac{1}{w^{2} }[/tex] (w + 2w)
= [tex]\frac{6w}{w^{2} }[/tex]
= [tex]\frac{6}{w }[/tex]
Since, material for the base costs $8 per [tex]m^{2}[/tex] . Material for the sides and lid costs $16 per [tex]m^{2}[/tex].
cost of sides are:
= {(2 * 2w * [tex]\frac{8}{w^{2} }[/tex]} + (2 * w * [tex]\frac{8}{w^{2} }[/tex])}(9)
= 176 / w
Cost of base is:
= [tex]2w^{2}[/tex] * 8
= 16[tex]w^{2}[/tex]
Now, Total cost = Cost of base + Cost of sides
C(w) = 16[tex]w^{2}[/tex] + 176 / w ...(1)
Differentiating (1) with respect to w, we get
C'(w) = 32w - [tex]\frac{176}{w^{2} }[/tex]
Differentiating again with respect to w
C''(w) = 32 + [tex]\frac{352}{w^{3} }[/tex]
For maxima or minima,
Put C'(w) = 0
32w - [tex]\frac{176}{w^{2} }[/tex] = 0
32[tex]w^{3}[/tex] - 176 = 0
w = 1.76
For w = 1.76.
C''(w) = positive
Hence, for width 1.76 m the cost is minimum.
Therefore, the minimum cost is
C(1.76) = 16[tex](1.76)^{2}[/tex] + 176 / 1.76
Minimum cost = $59.56
And the dimensions for which the cost is minimum is:
= 3.52m × 1.76m × 1.76m
Know more about Maxima and minima: -https://brainly.in/question/8699
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