 Lucille Davidson

2021-12-08

Find the absolute maximum and minimum values of $f\left(x,y\right)={x}^{2}+{y}^{2}-2x+1$ on the unit disc ${x}^{2}+{y}^{2}\le 1$. Marcus Herman

Step 1
Given function is: $f\left(x,y\right)={x}^{2}+{y}^{2}-2x+1$ and disc is: ${x}^{2}+{y}^{2}\le 1$
Step 2
Let,
$x=r\mathrm{cos}\left(0\right),y=r\mathrm{sin}\left(0\right)$
Then,
$f\left(r,0\right)={r}^{2}{\mathrm{cos}}^{2}\left(0\right)+{r}^{2}{\mathrm{sin}}^{2}\left(0\right)-2r\mathrm{cos}\left(0\right)+1$
$={r}^{2}-2r\mathrm{cos}\left(0\right)+1$
Step 3
Let,
$t=\mathrm{cos}\left(0\right)$
Then,

Taking partial derivatives,

${f}_{r}=2r-2t$
${f}_{t}=-2r$
Now, ${f}_{r}=0,{f}_{t}=0$ gives
r=0, t=0
Step 4
If r=0 only, then $x=r\mathrm{cos}\left(0\right)=0,y=r\mathrm{sin}\left(0\right)=0$
If t=0 only, then x=rt, $y=±r$
From this, at end point $\left(0,±1\right),\left(1,±1\right)$,
${f}_{max}=1+1=2$
${f}_{min}=1+1-2+1=1$
Step 5
Hence, the absolute maximum values is 2 and absolute minimum value is 1.

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