Question

If \[ a^2 x^4 + b^2 y^4 = c^6, \] then the maximum value of \( xy \) is:

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

Hint:

To find maximum values in constrained equations, use AM-GM or Lagrange multipliers.

Answer 1

The Correct Option is B

Step 1: Using AM-GM inequality For maximum \( xy \), apply the method of Lagrange multipliers or use the AM-GM inequality: \[ a^2 x^4 + b^2 y^4 \geq 2 \sqrt{a^2 x^4 b^2 y^4}. \] Substituting: \[ c^6 \geq 2 \sqrt{a^2 x^4 b^2 y^4}. \] Solving for \( xy \): \[ xy \leq \frac{c^3}{\sqrt{2ab}}. \]

Answer 2

Given the equation:
\[ a^2 x^4 + b^2 y^4 = c^6, \] find the maximum value of \( xy \).

Step 1: Let \( u = x^2 \) and \( v = y^2 \), so the equation becomes:
\[ a^2 u^2 + b^2 v^2 = c^6 \] We want to maximize:
\[ xy = \sqrt{uv} \]

Step 2: Use Lagrange multipliers to maximize \( f(u,v) = \sqrt{uv} \) subject to constraint:
\[ g(u,v) = a^2 u^2 + b^2 v^2 - c^6 = 0 \]

Step 3: Define Lagrangian:
\[ \mathcal{L} = \sqrt{uv} - \lambda (a^2 u^2 + b^2 v^2 - c^6) \]

Step 4: Calculate partial derivatives and set to zero:
\[ \frac{\partial \mathcal{L}}{\partial u} = \frac{1}{2} \frac{v^{1/2}}{u^{1/2}} - \lambda 2 a^2 u = 0 \] \[ \frac{\partial \mathcal{L}}{\partial v} = \frac{1}{2} \frac{u^{1/2}}{v^{1/2}} - \lambda 2 b^2 v = 0 \]

Step 5: From the first equation:
\[ \frac{v^{1/2}}{2 u^{1/2}} = 2 \lambda a^2 u \implies \frac{v^{1/2}}{u^{1/2}} = 4 \lambda a^2 u \]
From the second equation:
\[ \frac{u^{1/2}}{2 v^{1/2}} = 2 \lambda b^2 v \implies \frac{u^{1/2}}{v^{1/2}} = 4 \lambda b^2 v \]

Step 6: Multiply these two equations:
\[ \left( \frac{v^{1/2}}{u^{1/2}} \right) \times \left( \frac{u^{1/2}}{v^{1/2}} \right) = (4 \lambda a^2 u)(4 \lambda b^2 v) \] \[ 1 = 16 \lambda^2 a^2 b^2 u v \] \[ \lambda^2 = \frac{1}{16 a^2 b^2 u v} \]

Step 7: Divide the first equation by the second:
\[ \frac{\frac{v^{1/2}}{u^{1/2}}}{\frac{u^{1/2}}{v^{1/2}}} = \frac{4 \lambda a^2 u}{4 \lambda b^2 v} \implies \frac{v}{u} = \frac{a^2 u}{b^2 v} \implies b^2 v^2 = a^2 u^2 \] \[ \Rightarrow v = \frac{a}{b} u \]

Step 8: Substitute \( v = \frac{a}{b} u \) into the constraint:
\[ a^2 u^2 + b^2 \left( \frac{a}{b} u \right)^2 = c^6 \] \[ a^2 u^2 + a^2 u^2 = c^6 \] \[ 2 a^2 u^2 = c^6 \implies u^2 = \frac{c^6}{2 a^2} \implies u = \frac{c^3}{a \sqrt{2}} \]

Step 9: Find \( v \):
\[ v = \frac{a}{b} u = \frac{a}{b} \times \frac{c^3}{a \sqrt{2}} = \frac{c^3}{b \sqrt{2}} \]

Step 10: Finally, the maximum value of \( xy = \sqrt{uv} \):
\[ xy = \sqrt{uv} = \sqrt{ \frac{c^3}{a \sqrt{2}} \times \frac{c^3}{b \sqrt{2}} } = \sqrt{ \frac{c^6}{ab \times 2} } = \frac{c^3}{\sqrt{2ab}} \]

Therefore,
\[ \boxed{\frac{c^3}{\sqrt{2ab}}} \]