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Grade: 12

                        

The maximum possible value of (xv-yu) 2 over the surface given by the equations x 2 + y 2 ​ = 4 and u 2 +v 2 ​= 9 is ?

4 years ago

Answers : (2)

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 $\frac{x}{u} = -\frac{y}{v}$
4 years ago
Aditya Gupta
2069 Points
							
note that 36= (x^2+y^2)(u^2+v^2)= (xv – yu)^2 + (xu+yv)^2
so (xv – yu)^2= 36 – (xu+yv)^2
so max value occurs when (xu+yv)^2= 0
MAX value= 36 – 0
36
KINDLY APPROVE :))
4 months ago
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