Find the absolute maximum and minimum values of the function f(x,y)=2+2x+2y-x^2-y^2 on the triangular region in the first quadrant bounded by x=0,y=0,y=9-x.
This is the work I've done. I just need someone to verify that I've done it correctly. Is my methodology sound?
f(1,1) = 4
f(0,y)=2+2y-y^2 where 0≤y≤9; min=(0,9)=-61, max=(0,1)=3
Since this is a parabola that opens down, the max will occur when the derivative equals 0 (vertex of parabola).
2-2y=0 when y=1
Since the interval restricts the graph to be decreasing, the minimum will occur when y is the greatest on that interval. Hence,
note: "y" essentially acts as the x since I am testing this as a single variable function.
f(x,0)=2+2x-x^2, where 0≤x≤9; min=(9,0)-61, max=(1,0)=3
Since we are testing this as a single variable function, this graph takes the same shape as the previous one and has the same max and min.
f(x,-x+9)= -2x^2+18x-61, where 0≤x≤9; min = (9,0)-61 max=(4.5,4.5) = -20.5 (max at 4.5 by first derivative test)
The minimum will occur at one of the endpoints on the interval and f(0,9)= -61 and f(9,0)= -61 so these are both a min.
Since this is a parabola that opens downward, the max will occur when the derivative equals 0 (vertex of parabola).
-4x+18=0 when x=4.5
so max=(4.5,4.5) = -20.5
So comparing the extreme values on the boundary and the value of the critical point, it follows that:
Absolute Maximum = 4
Absolute Minimum = -61