Including the constraint, the first-order conditions for the solution of the problem are
L 1(x,y) = y −2λ =0, L 2(x,y) = x −λ =0, 2x +y =100 The first two equations imply that y = 2λ and x = λ. Soy = 2x. Inserting this into the constraint yields 2x +2x =100. So x =25 and y =50, implying that λ = x =25. This solution can be confirmed by the substitution method. From 2x + y = 100 we get y = 100−2x, so the problem is reduced to maximizing the unconstrained function h(x) = x(100−2x)=− 2x2 +100x. Since h(x) =− 4x +100 = 0 givesx = 25, and h(x) =−
Including the constraint, the first-order conditions for the solution of the problem areL 1(x,y) = y −2λ =0, L 2(x,y) = x −λ =0, 2x +y =100 The first two equations imply that y = 2λ and x = λ. Soy = 2x. Inserting this into the constraint yields 2x +2x =100. So x =25 and y =50, implying that λ = x =25. This solution can be confirmed by the substitution method. From 2x + y = 100 we get y = 100−2x, so the problem is reduced to maximizing the unconstrained function h(x) = x(100−2x)=− 2x2 +100x. Since h(x) =− 4x +100 = 0 givesx = 25, and h(x) =−
การแปล กรุณารอสักครู่..
