# Suppose f(x,y)=(x-y)(16-xy). Answer the following 1. Find the local maxima of

Suppose $$\displaystyle{f{{\left({x},{y}\right)}}}={\left({x}-{y}\right)}{\left({16}-{x}{y}\right)}$$. Answer the following
1. Find the local maxima of $$\displaystyle{f}$$.
2. Find the local minima of $$\displaystyle{f}$$.
3. Find the saddle points of $$\displaystyle{f}$$.

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lalilulelo2k3eq
$$\displaystyle{f{{\left({x},{y}\right)}}}={\left({x}-{y}\right)}{\left({16}-{x}{y}\right)}$$
$$\displaystyle{f}_{{x}}={\left({16}-{x}{y}\right)}+{\left({x}-{y}\right)}{\left(-{y}\right)}={16}-{x}{y}+{y}^{{2}}-{x}{y}$$
$$\displaystyle{f}_{{y}}={\left({x}{y}-{16}\right)}+{\left({x}-{y}\right)}{\left(-{x}\right)}={x}{y}-{16}-{x}^{{2}}+{x}{y}$$
$$\displaystyle{f}_{{x}}={0}\ \Rightarrow\ {y}^{{2}}-{2}{x}{y}-{16}={0};\ \Rightarrow{y}^{{2}}={x}^{{2}}$$
$$\displaystyle{f}_{{y}}={0}\ \Rightarrow\ {x}^{{2}}-{2}{x}{y}-{16}={0};\ \Rightarrow{y}=\pm{x}$$
$$\displaystyle{y}={x}\ \Rightarrow\ {x}^{{2}}-{2}{x}^{{2}}+{16}={0}\ \Rightarrow\ {x}^{{2}}={16}\ \Rightarrow\ {x}=\pm{4}$$
$$\displaystyle{y}=-{x}\ \Rightarrow\ {x}^{{2}}+{2}{x}^{{2}}+{16}={0}\ \Rightarrow\ {x}^{{2}}=-{\frac{{{16}}}{{{3}}}}$$ - not possible
$$\displaystyle{\left({x},{y}\right)}={\left(-{4},-{4}\right)},{\left({4},{4}\right)}$$
$$\displaystyle{f}_{{\times}}=-{2}{y};\ {f}_{{{y}{y}}}={2}{x};\ {f}_{{{x}{y}}}=-{2}{x}+{2}{y}$$
$$\displaystyle{D}=-{4}{x}{y}-{4}{\left({x}-{y}\right)}^{{2}}$$
$$\displaystyle{D}{\left({4},-{4}\right)}=-{4}{\left({16}\right)}=-{64}{ < }{0}$$
$$\displaystyle{D}{\left({4},{4}\right)}=-{4}{\left({16}\right)}=--{64}{ < }{0}$$
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psor32
nick1337
1) none

2) none

3) (-4,-4,0), (4,4,0)