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  4. The greatest value of (2 sin θ+3 cos θ+4)3 ⋅(6-2 sin θ-3 cos θ)2, as θ ∈ R, is

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Q. The greatest value of $(2 \sin \theta+3 \cos \theta+4)^{3} \cdot(6-2 \sin \theta-3 \cos \theta)^{2}$, as $\theta \in R$, is

Trigonometric Functions

Solution:

Clearly, $2 \sin \theta+3 \cos \theta+4 > 0 \forall \theta \in R$
and $6-2 \sin \theta-3 \cos \theta > 0 \forall x \in R$
Put. $2 \sin \theta+3 \cos \theta+4=x$ and $6-2 \sin \theta-3 \cos \theta=y$
Now, $x+y=10$
Also, $A.M \geq G.M$. (for positive numbers)
$\therefore \frac{\frac{x}{3}+\frac{x}{3}+\frac{x}{3}+\frac{y}{2}+\frac{y}{2}}{5} \geq\left(\frac{x^{3}}{27} \cdot \frac{y^{2}}{4}\right)^{\frac{1}{5}}$
$\therefore (2 \sin \theta+3 \cos \theta+4)^{3}(6-2 \sin \theta-3 \cos \theta)^{2}$
$\leq(32)(27)(4)=3456$