. By inspection, we can see that y cannot be 0 since y= 0 would yield 6=0. Case1) if x , y, z are not 0, we can solve for since we are not dividing by 0. We get Setting 1st = 3nd produces 1st= 2rd produces . But since Now substitute this into the constraint to get ,(there are 8 points) In this case, by going back to the optimization function, the maximum value is (when all the variables are positive) and the minimum is when x and z have the opposite signs and y is negative. Case 2) x= 0. In this case, since , z is also 0. ( We observed earlier that y cannot be 0). Going back to the constraint, . This case produces the f-values . Combining the two cases, we see that the max is 18, min is -18. Remember, dx ranges from the left equation to the right equation. On the left, On the right, . Thus the region looks like Using dy first, you must break up the region into two pieces: from 0 to 1, then from 1 to 2 since the equation changes at x = 1. It is given that We are seeking . First find R: . Differentiate implicitly, we get = Solving 4x-y=0, -x+2y=7 , we obtain x = 1, y=4. . (1.4) is a local minimum 50) a) b) Interior: fx= 2-2x=0,f y-=2-y=0: (1,1) Boundary: (0,1) 3 (1,0) 3 (9/2,9/2) -59/2 (0,0) 0
. By inspection, we can see that y cannot be 0 since y= 0 would yield 6=0. Case1) if x , y, z are not 0, we can solve for since we are not dividing by 0. We get Setting 1st = 3nd produces 1st= 2rd produces . But since Now substitute this into the constraint to get ,(there are 8 points) In this case, by going back to the optimization function, the maximum value is (when all the variables are positive) and the minimum is when x and z have the opposite signs and y is negative. Case 2) x= 0. In this case, since , z is also 0. ( We observed earlier that y cannot be 0). Going back to the constraint, . This case produces the f-values . Combining the two cases, we see that the max is 18, min is -18. Remember, dx ranges from the left equation to the right equation. On the left, On the right, . Thus the region looks like Using dy first, you must break up the region into two pieces: from 0 to 1, then from 1 to 2 since the equation changes at x = 1. It is given that We are seeking . First find R: . Differentiate implicitly, we get = Solving 4x-y=0, -x+2y=7 , we obtain x = 1, y=4. . (1.4) is a local minimum 50) a) b) Interior: fx= 2-2x=0,f y-=2-y=0: (1,1) Boundary: (0,1) 3 (1,0) 3 (9/2,9/2) -59/2 (0,0) 0