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# Finding maxima and minima of a mathematical function

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Find the local maximum and minimum values and the saddle points of f(x,y) = 2xy + x-y.

https://brainmass.com/math/calculus-and-analysis/finding-maxima-and-minima-of-a-mathematical-function-18497

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A function of two variables has a local maximum at (a,b)
<br>if f(x,y) =&lt; f(a,b) when (x,y) is near (a,b)
<br>
<br>The number f(a,b) is called a local maximum value.
<br>
<br>If f(x,y) &gt;= f(a,b) when (x,y) is near (a,b) then f(a,b) is a local minimum value.
<br>
<br>
<br>our function is,
<br>
<br>f(x,y) = 2xy + x -y
<br>
<br>let me use fx = partial derivative with respect to x and fy for y
<br>
<br>fx = 2 y + 1 -----(1)
<br>
<br>fy = 2x - 1 -----(2)
<br>
<br>to find the crical point, solve the equations,
<br>
<br>2y+ 1 = 0 and 2x-1 = 0
<br>
<br>2y+1=0 ====&gt; y = -1/2
<br>
<br>2x-1=0 ====&gt; x = 1/2
<br>
<br>Therefore, the critical point is (1/2, -1/2)
<br>
<br>Now to the second derivative test
<br>
<br>D = D(a,b) = fxx (a,b) * fyy(a,b)- [fxy (a,b)]^2
<br>
<br>fxx(a,b) = 0 (differentiate eq 1 with respect to x again)
<br>fyy(a,b) = 0
<br>
<br>fxy = 2 (differentiate eq 1 with respect to y)
<br>
<br>at(1/2, -1/2)
<br>
<br>D = 0 - 2 = -2
<br>
<br>Now refer to the figure, (1/2, -1/2) is a saddle point

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