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Given $f(x,y) = x^2+y^2-14x-20y$ and restrictions $x≥0,\; 0≤y≤42 \;\text{and}\; y≥x$, I need to find the max and min.

By finding partial derivatives and setting them to $0$, I get the min to be -149. However, I'm having trouble finding the max.

Lauralolo
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  • We need to identify the boundary of our region. This is the boundary of the triangle with vertices $(0,0)$, $(42,42)$, and $(0,42)$. The candidates for max/min are the places inside the triangle at which the partials are equal to $0$, together with the boundary points. Note that the partial derivatives are $0$ at $(7,10)$. One also needs to explore the boundary. – André Nicolas Mar 01 '15 at 00:09

1 Answers1

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i am wondering if we can do this. $$f = x^2 + y^2 - 14x - 20 y = (x-7)^ 2 + (y - 10)^2-149 $$ and using the fact that $f = const$ is a circle. so the biggest value of $f$ is obtained for the largest radius, which occurs at the corner $x = 42, y = 42.$

therefore the maximum value of $f$ is $$ f = x^2 + y^2 - 14x - 20y = (42-7)^ 2 + (42 - 10)^2-149 = 2100$$

abel
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