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Question
Divide the number 20 into two parts such that sum of their squares is minimum.
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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