Question

If $a^2x^4+b^2y^4=c^6$, then maximum value of $xy$ is

• A. $\frac{c^2}{\sqrt{ab}}$
• B. $\frac{c^3}{ab}$
• C. $\frac{c^3}{\sqrt{2ab}}$
• D. $\frac{c^3}{2ab}$

Answer 1

Answer: (C)

$\frac{c^3}{\sqrt{2ab}}$

Answer 2

To find the maximum value of $xy$ given the equation $a^2x^4+b^2y^4=c^6$, we can use the AM-GM inequality.

1. Apply AM-GM Inequality: The AM-GM inequality states that for two non-negative numbers $A$ and $B$, their arithmetic mean is greater than or equal to their geometric mean: $\frac{A+B}{2} \ge \sqrt{AB}$ Equality holds when $A=B$.

Let $A = a^2x^4$ and $B = b^2y^4$. Since $x^4$ and $y^4$ are always non-negative, and $a^2, b^2$ are also non-negative, $A$ and $B$ are non-negative. Substitute $A$ and $B$ into the AM-GM inequality: $\frac{a^2x^4+b^2y^4}{2} \ge \sqrt{a^2x^4 \cdot b^2y^4}$ We are given $a^2x^4+b^2y^4=c^6$. Substitute this into the left side: $\frac{c^6}{2} \ge \sqrt{a^2b^2x^4y^4}$ Simplify the right side: $\frac{c^6}{2} \ge \sqrt{(ab)^2 (x^2y^2)^2}$ $\frac{c^6}{2} \ge |abx^2y^2|$ Since $x^2y^2$ is non-negative, and for the maximum value of $xy$ to be positive, we assume $ab > 0$. Thus, $|abx^2y^2| = abx^2y^2$. $\frac{c^6}{2} \ge abx^2y^2$ Rearrange the inequality to find the upper bound for $x^2y^2$: $x^2y^2 \le \frac{c^6}{2ab}$ To maximize $xy$, we need to maximize $x^2y^2$. The maximum value of $x^2y^2$ is $\frac{c^6}{2ab}$.

2. Find the maximum value of $xy$: Take the square root of the maximum value of $x^2y^2$: $xy = \sqrt{\frac{c^6}{2ab}}$ $xy = \frac{\sqrt{c^6}}{\sqrt{2ab}}$ $xy = \frac{c^3}{\sqrt{2ab}}$

3. Condition for Maximum Value: The maximum value occurs when the equality in the AM-GM inequality holds, which means $A=B$: $a^2x^4 = b^2y^4$ Substitute this condition back into the original equation $a^2x^4+b^2y^4=c^6$: $a^2x^4 + a^2x^4 = c^6$ $2a^2x^4 = c^6 \implies x^4 = \frac{c^6}{2a^2}$ Similarly, $b^2y^4 + b^2y^4 = c^6$ $2b^2y^4 = c^6 \implies y^4 = \frac{c^6}{2b^2}$ Now, calculate $x^2y^2$: $x^2 = \sqrt{\frac{c^6}{2a^2}} = \frac{c^3}{\sqrt{2}a}$ $y^2 = \sqrt{\frac{c^6}{2b^2}} = \frac{c^3}{\sqrt{2}b}$ (We take the positive square roots since $x^2$ and $y^2$ are non-negative.) Then, $x^2y^2 = \left(\frac{c^3}{\sqrt{2}a}\right) \left(\frac{c^3}{\sqrt{2}b}\right) = \frac{c^6}{2ab}$ Finally, $xy = \sqrt{\frac{c^6}{2ab}} = \frac{c^3}{\sqrt{2ab}}$.

This confirms the maximum value derived from the AM-GM inequality.