Can someone help with this question please?
1 answer:
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Answer:
Dimensions are 2 m by 1 meter by 1 meter,
Minimum cost is $ 18.
Step-by-step explanation:
Let w be the width ( in meters ) of the container,
Since, the length is twice of the width,
So, length of the container = 2w,
Now, if h be the height of the container,
Volume = length × width × height
2 = 2w × w × h
1 = w² × h
Since, the area of the base = l × w = 2w × w = 2w²,
Area of the lid = l × w = 2w²,
While the area of the sides = 2hw + 2hl
= 2h( w + l)
Since, Material for the base costs $1 per m². Material for the sides and lid costs $2 per m²,
So, the total cost,
Differentiating with respect to w,
Again differentiating with respect to w,
For maxima or minima,
C'(w) = 0
For w = 1, C''(w) = positive,
Hence, for width 1 m the cost is minimum,
Therefore, the minimum cost is C(1) = 6(1)²+12 = $ 18,
And, the dimension for which the cost is minimum is,
2 m by 1 meter by 1 meter.