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Question
Divide 20 into two ports, so that their product is maximum.
Solution
Let one part of 20 be x.
∴ The other part is (20 – x)
∴ Product = x.(20 – x)
Which has to be maximized.
∴ f(x) = x.(20 – x)
= 20x – x2
∴ f'(x) = 20 – 2x
f"(x) = – 2 < 0
Let f'(x) = 0
∴ 20 – 2x = 0
⇒ 2x = 20
⇒ x = 10
And f"(x) = – 2 < 0
∴ By 2nd derivative test, f is maximum at x = 10
∴ 20 – x = 20 – 10 = 10.
∴ The required parts of 20 are 10 and 10.
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Complete the following activity to divide 84 into two parts such that the product of one part and square of the other is maximum.
Solution: Let one part be x. Then the other part is 84 - x
Letf (x) = x2 (84 - x) = 84x2 - x3
∴ f'(x) = `square`
and f''(x) = `square`
For extreme values, f'(x) = 0
∴ x = `square "or" square`
f(x) attains maximum at x = `square`
Hence, the two parts of 84 are 56 and 28.
