Given $f(x,y) = x+ y$ and $g(x,y) = x^2 + y^2$ as well as the knowledge that $f$ has its biggest value at a point $(a,b)$ on the domain $g(x,y)=2$, what is $\bigtriangledown g(a,b)$?
Not quitre sure how to proceed.
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Solve x^2+y^2=2 for y. Calculate the gradient of g. Use solution with f ... – Loreno Heer Nov 06 '14 at 13:16
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$(a,b)$ should be on the domain $g(x,y)=x^2+y^2=2$, so first express say $y$ in terms of $x$ as $y=\sqrt{2-x^2}$ and substitute it in $f(x,y)$: $$f(x,y)=x+\sqrt{2-x^2}$$ Now maximize this 1-variable function with respect to $x$: $$\frac{d f}{d x}=1+\frac{1}{2}(-2x)(2-x^2)^{\frac{-1}{2}}=0$$ which is going to give $x=1$. Therefore $y=1$ and $(a,b)=(1,1)$. Considering this:
$$\bigtriangledown g(a,b)=(2,2)$$
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