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Question
Divide 20 into two ports, so that their product is maximum.
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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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