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The maximum possible value of (xv-yu)2 over the surface given by the equations x2 + y2​ = 4 andu2+v2​= 9 is ?

jagdish singh singh
173 Points
4 years ago
$\hspace{-0.7 cm} Using Vector Inequality: Let \vec{a}=x\hat{i}+y\hat{j} and \vec{b}=v\hat{i}-u\hat{j}\\\\ So \vec{a}\cdot \vec{b} = |\vec{a}||\vec{b}|\cos \phi\leq |\vec{a}||\vec{b}|\\\\ So (\vec{a}\cdot \vec{b})^2\leq |\vec{a}|^2|\vec{b}|^2\\\\ So (xv-yu)^2 \leq (x^2+y^2)(u^2+v^2) = 4\cdot 9=36\\\\ and equality hold when \frac{x}{u} = -\frac{y}{v}$
2075 Points
11 months 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 :))