Consider the function fxyxy 6y 36x216 on the region on or...

consider the function f(x,y)=xy-6y-36x+216 on the region on or above y=x^(2) and on or below y=39. Find the absolute minimum value: Find the points at which the absolute minimum value is attained. List your answer as points in the form (a,b). Find the absolute maximum value: Find the points at which the absolute maximum value is attained. List your answer as points in the form (a,b).

Transcript

00:01 Okay so we have this function f of x y equals x y minus 6 y minus 36 x plus 216 x and what we need to do is we need to first find local min and maxes so let's find local min and max and then we'll worry about the constraints so local min and max what do we need to do is we need to find grad f equals zero solve this equation so what is grad f so grad f you take first the partial derivative respect to x so you get y minus 36 and then you take a partial derivative respect to y so you get x minus 6 and you set this equal to zero so this means that y is equal to 36 x is equal to 6 is a critical point okay so this is one possible value 6 36 and note that it's in the the feasible region so it is below y equals 40 so we have the constraints they have to be so you have y equals x squared and y equals 40 here and x 36 is there so it's allowed it's in the possible region 6 36 so this is allowed possible min and max and then now that we did the local points now i need to think about what's happening on the boundaries of this region so what is the boundary of this region the boundary of the region first boundary is y equals 40 so we need to write out our function so we write f of x 40 so this becomes 40 x minus 6 times 40 is 240 minus 36 x plus 216 so 40 x minus 36 x is 4 x so you get 4 x and then here you get minus 24 okay and then you have to find the max you have to maximize this function along the possible x values so as we can see the max is when x is the largest it can be and what is the largest x that's possible the largest x is where y equals x squared intersects with y equals 40 so it's at square root of 40 so the this is at a maximum or a minimum okay and what's the minimum the minimum is when you're at the other end of this feasible region which is minus square root of 40 why is this okay so other possible points are when x is equal to plus or minus square root of 40 and y is equal to 40 so that's at this first boundary now what about the second boundary the second boundary is the parabola this is the graph y is equal to x squared so let's plug it in so we have f of x x squared so if you plug this in you get x cubed minus 6 x squared minus 36 x minus 216 or plus 216 okay so and how do you maximize this or minimize this well we need to take the derivative…

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