If $a + 3 b + 2 c = 12$ , then the maximum value of $6 b c (1 + a) + a (3 b + 2 c)$ is

112

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

115

No worries! We‘ve got your back. Try BYJU‘S free classes today!

102

No worries! We‘ve got your back. Try BYJU‘S free classes today!

92

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $112$
Consider the positive numbers $(1 + a), (1 + 3 b), (1 + 2 c)$
The AM $\geq$ GM.
So, $\frac{3 + a + 3 b + 2 c}{3} \geq {(1 + a) (1 + 3 b) (1 + 2 c)}^{\frac{1}{3}} \Rightarrow 5 \geq {(1 + 12 + 6 b c (1 + a) + a (3 b + 2 c)}^{\frac{1}{3}} \Rightarrow 5^{3} \geq 13 + 6 b c (1 + a) + a (3 b + 2 c) \Rightarrow 112 \geq 6 b c (1 + a) + a (3 b + 2 c)$
Hence, option A is the correct answer.

Suggest Corrections

Similar questions

If a+b+c = 13, then what is the maximum value of

(a - 3)( b - 2)(c + 1)

\to a, \to b, \to c

{∣ ∣ ∣ 2 \to a - 3 \to b ∣ ∣ ∣}^{2} + {∣ ∣ ∣ 2 \to b - 3 \to c ∣ ∣ ∣}^{2} + {∣ ∣ 2 \to c - 3 \to a ∣ ∣}^{2}

Please Use the Components of Proportion like Componendo and dividends etc

If (a+3b+2c+6d)(a-3b-2c+6d)=(a+3b-2c-6d)(a-3b-2c+2c-6d)

Prove that a:b::c:d

3^{a} + 3^{b} = 756, 7^{a} + 2^{c} = 375, 5^{a} + 3 = 128

a + b + c

Join BYJU'S Learning Program