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Salsk061 [2.6K]
3 years ago
9

Find the maximum and minimum values of the function f(x,y)=2x2+3y2−4x−5 on the domain x2+y2≤100. The maximum value of f(x,y) is:

Mathematics
1 answer:
ryzh [129]3 years ago
4 0

First find the critical points of <em>f</em> :

f(x,y)=2x^2+3y^2-4x-5=2(x-1)^2+3y^2-7

\dfrac{\partial f}{\partial x}=2(x-1)=0\implies x=1

\dfrac{\partial f}{\partial y}=6y=0\implies y=0

so the point (1, 0) is the only critical point, at which we have

f(1,0)=-7

Next check for critical points along the boundary, which can be found by converting to polar coordinates:

f(x,y)=f(10\cos t,10\sin t)=g(t)=295-40\cos t-100\cos^2t

Find the critical points of <em>g</em> :

\dfrac{\mathrm dg}{\mathrm dt}=40\sin t+200\sin t\cos t=40\sin t(1+5\cos t)=0

\implies\sin t=0\text{ OR }1+5\cos t=0

\implies t=n\pi\text{ OR }t=\cos^{-1}\left(-\dfrac15\right)+2n\pi\text{ OR }t=-\cos^{-1}\left(-\dfrac15\right)+2n\pi

where <em>n</em> is any integer. We get 4 critical points in the interval [0, 2π) at

t=0\implies f(10,0)=155

t=\cos^{-1}\left(-\dfrac15\right)\implies f(-2,4\sqrt6)=299

t=\pi\implies f(-10,0)=235

t=2\pi-\cos^{-1}\left(-\dfrac15\right)\implies f(-2,-4\sqrt6)=299

So <em>f</em> has a minimum of -7 and a maximum of 299.

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