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Question
Divide the number 30 into two parts such that their product is maximum.
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Solution
Let the first part of 30 – x.
∴ Their product
= x(30 – x)
= 30x – x2
= f'(x) ......(Say)
∴ f'(x) = `"d"/("d"x)(30x - x^2)`
= 30 × 1 − 2x
= 30 − 2x
and f'(x) = `"d"/("d"x)(30 - 2x)`
= 0 – 2 × 1
= – 2
The root of the equation f'(x) = 0
i.e. 30 – 2x = 0 is x = 15
and f'(15) = – 2 < 0
∴ By the second derivative test, f is maximum at x = 15.
Hence, the required parts of 30 are 15 and 15.
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