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Anyone good at max and min probs? (1 Viewer)

Petinga

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A rectangular block, the length of whose base is twice the width, has a toatl surface area of 300cm squared. Find the dimensions of the block if it is of maximum volume?

Need help desperately
 

shafqat

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Petinga said:
A rectangular block, the length of whose base is twice the width, has a toatl surface area of 300cm squared. Find the dimensions of the block if it is of maximum volume?

Need help desperately
Let width be x, length 2x, and the other dimension y. Use the surface area: 4x^2 + 6xy = 300. Solve for y. Then find an expression for the volume by multiplying the 3 dimensions. Maximimum by differentiation.
 
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shafqat said:
Let width be x, length 2x, and the other dimension y. Use the surface area: 4x^2 + 6xy = 300. Solve for y. Then find an expression for the volume by multiplying the 3 dimensions. Maximimum by differentiation.
Finishing this off...

4x^2 + 6xy = 300

6xy = 300 - 4x^2

y = 50/x - (2/3)x

Volume (V) = x * 2x * y = 2(x^2)*y = 100x - (4/3)x^3

V' = 100 - 4x^2

When V' = 0, 4x^2 = 100

x^2 = 25

x = 5 (x is a length, and lengths are positive)

V'' = -8x, which is negative at x = 5.

Therefore x = 5 is an absolute maximum.

Therefore the maximum volume is when dimensions are 5cm, 10cm and 20/3 cm
 

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