Correct Answer - Option 1 : 100

**CONCEPT**:

Second Derivative Test: Let f be a function defined on an interval I.

- Calculate f’(x)
- Solve f’(x) = 0 and find the roots of f'(x) = 0. Suppose x = c is the root of f’(x) = 0.
- Calculate f’’(x) and put x = c to get the value of f’’(c).
- If f’’(c) < 0 then x = c is a point of local maxima.
- If f’’(c) > 0 then x = c is a point of local minima.
- If f’’(c) = 0 then we need to use the first derivative test.

Note:

Maxima and Minima on a closed Interval:

Let f(x) be a given function defined on [a, b].Let the local minimum value of f(x) be m and let the local maximum value of f(x) be M. Then

Minimum value of f(x) on [a, b] is the smallest of m, f(a) and f(b)

Maximum value of f(x) on [a, b] is the greatest of M, f(a) and f(b)

**CALCULATION**:

Given: x + y = 20 and P = xy

⇒ P = x ⋅ (20 - x) = 20x - x^{2}

⇒ P(x) = 20x - x^{2}

⇒ P'(x) = 20 - 2x

If P'(x) = 0 ⇒ 20 - 2x = 0 ⇒ x = 10

By substituting x = 10 in the equation x + y = 20 we get, y = 10

⇒ P''(x) = - 2

⇒ P''(x = 10) = - 2 < 0

So, x = 10 is the point of maxima

So, the maximum value of P = xy = 10 × 10 = 100

Hence, option 1 is correct.