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Question
Divide the number 20 into two parts such that sum of their squares is minimum.
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Solution
Let the first part of 20 be x.
Then the second part is 20 – x.
∴ Sum of their squares = x2 + (20 – x)2 = f(x) ...(Say)
∴ f'(x) = `d/dx[x^2 + (20 - x)^2]`
= `2x + 2(20 - x) * d/dx(20 - x)`
= 2x + 2(20 – x) × (0 – 1)
= 2x – 40 + 2x
= 4x – 40
and f"(x) = `d/dx(4x - 40)`
= 4 × 1 – 0
= 4
The root of the equation f'(x) = 0, i.e., 4x – 40 = 0 is x = 10 and f"(10) = 4 > 0.
∴ By the second derivative test, f is minimum at x = 10.
Hence, the required parts of 20 are 10 and 10.
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