Question

minimize f=x^2+y^2 with x+2y=5

Answer 1

you wanna use calculus?

if so, single variable calculus?

if so, use \(x+2y = 5\) to replace x or y in \(f=f(x,y)\) so that you have \(f=f(x)\) or \(f=f(y)\)......as opposed to the current \(f=f(x,y)\)

and then do a derivative \(\dfrac{df}{dx}\) or \(\dfrac{df}{dy}\) to find turning points....and then check if min or max.

Answer 2

Another suggestion via Lagrange multipliers. With \(f(x,y)=x^2+y^2\) and \(g(x,y)=x+2y-5\), the Lagrangian is
\[\mathcal{L}(x,y,\lambda)=x^2+y^2+\lambda(x+2y-5)\]Take your partial dertivatives and set equal to \(0\):
\[\left.\begin{array}{l}
\dfrac{\partial\mathcal{L}}{\partial x}=2x+\lambda\\[1ex]
\dfrac{\partial\mathcal{L}}{\partial y}=2y+2\lambda\\[1ex]
\dfrac{\partial\mathcal{L}}{\partial \lambda}=x+2y-5
\end{array}\right\}=0\]From here it's just a matter of solving for \(\lambda\) and you'll get the minimum of \(f\) right away.

Answer 3

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