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2. A silo (base not included) is to be constructed in the form of a cylinder surmounted by a hemisphere. The cost of construction per square unit

of surface area is twice as great for the for the hemisphere as it is for the cylindrical side wall. Determine the dimensions to be used if the volume is fixed and the cost of construction is to be kept to a minimum. Neglect the thickness of the silo and waste in construction.​


Sagot :

An extremum is a point where the function has its highest or lowest value, and at which the slope is zero.

The dimensions to be used if the volume is fixed and the cost of construction is to be kept to a minimum is as follows;

  • Height of cylinder = 2 × Radius of cylinder

Reason:

The given parameters are;

Form of silo = Cylinder surmounted by a hemisphere

Cost of construction of hemisphere per square unit = 2 × The cost of of construction of the cylindrical side wall

Required:

The dimensions to be used if the volume is fixed and the cost of construction is to be kept to a minimum

Solution:

The fixed volume of the silo, V, can be expressed as follows;

[tex]V = \pi \cdot r^2 \cdot h + \dfrac{2}{3} \cdot \pi \cdot r^3[/tex]

  • [tex]h = \dfrac{V - \dfrac{2}{3} \cdot \pi \cdot r^3 }{\pi \cdot r^2 }[/tex]

Where;

h = The height of the cylinder

r = The radius of the cylinder

The surface area is, Surface Area = 2·π·r·h + 2·π·r²

The cost, C = 2·π·r·h + 2×2·π·r² = 2·π·r·h + 4·π·r²

Therefore;

[tex]C = 2 \times \pi \times r\times \dfrac{V - \dfrac{2}{3} \cdot \pi \cdot r^3 }{\pi \cdot r^2 } + 4 \cdot \pi \cdot r^2 = \dfrac{8 \cdot r^3 \cdot \pi + 6 \cdot V}{3 \cdot r}[/tex]

  • [tex]C = \dfrac{8 \cdot r^3 \cdot \pi + 6 \cdot V}{3 \cdot r}[/tex]

At the minimum value, we have;

  • [tex]\dfrac{dC}{dr} =0 = \dfrac{d}{dr} \left(\dfrac{8 \cdot r^3 \cdot \pi + 6 \cdot V}{3 \cdot r} \right) = \dfrac{16 \cdot r^3 \cdot \pi - 6 \cdot V}{3 \cdot r^2}[/tex]

Which gives;

16·π·r³ = 6·V

  • [tex]V = \dfrac{16 \cdot \pi \cdot r^3}{6} = \dfrac{8 \cdot \pi \cdot r^3}{3}[/tex]

Which gives;

[tex]h = \dfrac{\dfrac{8 \cdot \pi \cdot r^3}{3} - \dfrac{2}{3} \cdot \pi \cdot r^3 }{\pi \cdot r^2 } = 2 \cdot r[/tex]

h = 2·r

The height of the silo, h = 2 × The radius, 2

Therefore, the dimensions to be used if the volume is fixed and the cost is to be kept to a minimum is, height, h = 2 times the radius

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