MATH0801 Final: s) Opt Solutions
Document Summary
It follows that a=2 x 12 x 2 =24 x 2 x 3 . y x and hence da dx. Plugging this into the constraint yields y=8 . The maximum area is a=32 units2. y x x. It follows that a= x2 4 x 32000 x 2. =x 2 128000 x and hence da dx. The critical points are x=0 and x=40 . Thus, the dimensions of the box are 40 40 20 . The only positive critical point is x=1 , so the volume is. Let the two numbers be x and y. It follows that p= 200 y y=200 y y 2 and taking the derivative gives dp dy. Plugging this into the constraint gives x=100 . We can minimize d by minimizing d *= x 3 2 y 1 2. Hence the point on the line 6 x y=9 closest to 3,1 is 45.
