The base and sides will cost $.01 per cm2 to produce but the top, which is plastic and resealable, will cost $.02 per cm2 to produce. What s

Question

The base and sides will cost $.01 per cm2 to produce but the top, which is plastic and resealable, will cost $.02 per cm2 to produce. What should the dimensions be to minimize cost?

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Allison 2 weeks 2021-09-13T21:23:19+00:00 1 Answer 0

Answers ( )

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    2021-09-13T21:24:59+00:00

    Here is the full question

    A snack food company wishes to have a cylinder package for it’s almond and cashew mix. The cylinder must contain 120 cm³ worth of product. The base and sides will cost $.01 per cm2 to produce but the top, which is plastic and resealable, will cost $.02 per cm2 to produce. What should the dimensions be to minimize cost?

    Answer:

    The radius and height are both dimension in the cylinder; in order to minimize the cost

    radius = 2.515 cm

    height = 18.93 cm

    Step-by-step explanation:

    We denote the radius of the cylinder to be = r

    and the height of the cylinder = h

    The volume of a cylinder is known to be = πr²h

    Also, from the question; we are also told that the cylinder contains 120 πcm³

    i.e πr²h = 120π

    Dividing both sides with π; we have:

    r²h = 120

    h = \frac{120}{r^2}

    The base and sides will cost $.01 per cm² to produce

    Total cost of the base and side c_1 = 0.01 ( πr² + 2πrh)

    but the top, which is plastic and resealable, will cost $.02 per cm² to produce.

    i.e

    cost of the top cylinder c_2 = 0.02 ( πr²)

    Overall Total cost = c_1 + c_2

    = 0.01 ( πr² + 2πrh) + 0.02 ( πr²)

    = 0.01 πr² + 0.02 πrh + 0.02 πr²

    = 0.03 \pi r^2 + 0.02 \pi r (\frac{120}{r^2} )

    = 0.03 \pi r^2 + 2.4 \frac{\pi}{r}

    Taking the differentiation to find the radius dimension to minimize cost; we have:

    \frac{dc}{dr} =00.06 \pi r^2 - \frac{2.4 \pi}{r^2} =0

    0.06 \pi r^2 = \frac{2.4 \pi}{r^2}

    r^4 = \frac{2.4 \pi}{0.06 \pi}

    r^4 = 40

    r = \sqrt[4]{40}

    r= 2.515 cm

    However, \frac{d^2c}{dr^2}=0.12 \pi r + \frac{4.8 \pi}{r^3}

    \frac{d^2c}{dr^2}|__{r= 2.515}} =0.12 \pi (2.515) + \frac{4.8 \pi}{(2.515)^3} >0

    Therefore; we can say that the cost is minimum at r = 2.515 since it is positive.

    To determine the height ; we have:

    h = \frac{120}{r^2} \\h = \frac{120}{(2.515)^2}

    h = \frac{120}{6.34}

    h = 18.93 cm

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