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Grade: 12
        
The maximum possible value of (xv-yu)2 over the surface given by the equations x2 + y2​ = 4 and
u2+v2​= 9 is ?
3 years ago

Answers : (1)

jagdish singh singh
173 Points
							
\hspace{-0.7 cm}$ Using Vector Inequality: Let $\vec{a}=x\hat{i}+y\hat{j}$ and $\vec{b}=v\hat{i}-u\hat{j}$\\\\ So $\vec{a}\cdot \vec{b} = |\vec{a}||\vec{b}|\cos \phi\leq |\vec{a}||\vec{b}|$\\\\ So $(\vec{a}\cdot \vec{b})^2\leq |\vec{a}|^2|\vec{b}|^2$\\\\ So $(xv-yu)^2 \leq (x^2+y^2)(u^2+v^2) = 4\cdot 9=36$\\\\ and equality hold when $\frac{x}{u} = -\frac{y}{v}$
3 years ago
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