If a + 2b + 3c = 6,then the greatest value of | vedclass

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(A) Given the constraint $a + 2b + 3c = 6$,where $a, b, c > 0$.We want to maximize $abc^2$. We can rewrite the expression as $a(2b)(\frac{3c}{2})(\fra

Vedclass · Accepted Answer

$\frac{9}{8}$

Vedclass · Answer

(A) Given the constraint $a + 2b + 3c = 6$,where $a, b, c > 0$.We want to maximize $abc^2$. We can rewrite the expression as $a(2b)(\frac{3c}{2})(\frac{3c}{2}) = \frac{9}{4}abc^2$.Using the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality for the four terms $a, 2b, \frac{3c}{2}, \frac{3c}{2}$:$\frac{a + 2b + \frac{3c}{2} + \frac{3c}{2}}{4} \geq \sqrt[4]{a \cdot 2b \cdot \frac{3c}{2} \cdot \frac{3c}{2}}$Substituting the sum $a + 2b + 3c = 6$:$\frac{6}{4} \geq \sqrt[4]{a \cdot 2b \cdot \frac{9c^2}{4}}$$\frac{3}{2} \geq \sqrt[4]{\frac{18}{4} abc^2}$$\frac{3}{2} \geq \sqrt[4]{\frac{9}{2} abc^2}$Raising both sides to the power of $4$:$(\frac{3}{2})^4 \geq \frac{9}{2} abc^2$$\frac{81}{16} \geq \frac{9}{2} abc^2$$abc^2 \leq \frac{81}{16} \cdot \frac{2}{9} = \frac{9}{8}$.Thus,the maximum value is $\frac{9}{8}$.

If $a + 2b + 3c = 6$,then the greatest value of $abc^2$ is (where $a, b, c$ are positive real numbers).

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