The maximum value of the expression, $f (x, y, z) = x y z (4 d - a x - b y - c z)$ where a, b, c, d are positive constants, x, y, z are positive variables and $4 d - a x - b y - c z > 0$ is

\frac{a b c}{d}

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\frac{d^{2}}{a b c}

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\frac{a b c}{d^{3}}

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\frac{d^{4}}{a b c}

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Solution

The correct option is D $\frac{d^{4}}{a b c}$
$\frac{d f (x, y, z)}{d x} = (4 d y z - 2 a x y z - b y^{2} z - c y z^{2}) = 0 4 d - a x = (a x + b y + c z) (1) f^{''} (x, y, z) = (- 2 a y z) (m a x i m u m) 4 d - b y = (a x + b y + c z) (2) 4 d - c z = (a x + b y + c z) (3)$

Adding $(1), (2)$ and $(3)$
$12 d - (a x + b y + c z) = 3 (a x + b y + c z) 3 d = (a x + b y + c z) f (x, y, z) = (x y z) (4 d - 3 d) = (x y z) d f (x, x, x) = x^{3} (a + b + c) x = x^{4} (a + b + c) 3 d = (a + b + c) x a + b + c \geq 3 {(a b c)}^{1 / 3} x_{m a x} = \frac{3 d}{3 {(a b c)}^{1 / 3}}, y_{m a x} = \frac{3 d}{3 {(a b c)}^{1 / 3}}, z_{m a x} = \frac{3 d}{3 {(a b c)}^{1 / 3}} f (x, y, z) = (x y z) d = \frac{d^{4}}{a b c}$

Suggest Corrections

Similar questions

f (x, y, z) = x y z (4 d - a x - b y - c z)

4 d - a x - b y - c z > 0

a \neq b \neq c \neq 1; a x + y + z = 0

x + b y + z = 0, x + y +

= 0

a + b + c -

= \dots \dots \dots \dots \dots

x = b y + c z + d u, y = c z + a x + d u, z = a x + b y + d u, u = a x + b y + c z .

\frac{a}{1 + a} + \frac{b}{1 + b} + \frac{c}{1 + c} + \frac{d}{1 + d}

a^{x} = b^{y} = c^{z} = a b c

\frac{1}{x} + \frac{1}{y} + \frac{1}{z} =

a^{x} = b^{y} = c^{z} =

\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = . . . . . . .

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