If 2x + y = 6 then the maximum value of x^2 y is : Option 1: 27/4 Option 2: 27 Option 3: 8 Option 4: 18

Answer: 8. Solution: From 2x + y = 6, we get y = 6 - 2x. Let f(x) = x^2 (6 - 2x) = 6x^2 - 2x^3. f'(x) = 12x - 6x^2 = 6x(2 - x) = 0 x = 0 or x = 2. f''(x) = 12 - 12x; at x = 2, f''(2) = -12 < 0 (max). f(2) = 4(6 - 4) = 8.

Calculus

Application of Derivatives

Medium

If $2x + y = 6$ then the maximum value of $x^2 y$ is :

$\frac{27}{4}$

✅ Correct Option: 3

From $2x + y = 6$ , we get $y = 6 - 2x$ .

Let $f(x) = x^2 (6 - 2x) = 6x^2 - 2x^3$ .

$f'(x) = 12x - 6x^2 = 6x(2 - x) = 0 \Rightarrow x = 0$ or $x = 2$ .

$f''(x) = 12 - 12x$ ; at $x = 2$ , $f''(2) = -12 < 0$ (max).

$f(2) = 4(6 - 4) = 8$ .

2025: 15 May Shift 2

Match List-I with List-II

The function $f(x) = (x - 1)(x + 1)^2$ has

List-I	List-II
(A) A local maxima at $x =$ ____	(I) $\frac{1}{3}$
(B) A local minima at $x =$ ____	(II) 0
(C) The local minimum value of $f(x) =$ ____	(III) -1
(D) The local maximum value of $f(x) =$ ____	(IV) $-\frac{32}{27}$

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