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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 ?

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

Grade:12

2 Answers

jagdish singh singh
173 Points
5 years ago
\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 $\frac{x}{u} = -\frac{y}{v}$
Aditya Gupta
2081 Points
2 years ago
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 :))

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