MAT133Y1 Lecture Notes - Lecture 19: Paq, Lagrange Multiplier, Constrained Optimization
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MAT133Y1 Full Course Notes
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Mat133y1y - lecture 19 - chapter 17 (continued): A thermometer moves the xy plane so that its coordinates at time t are x9t) = t3 and y(t) = t2. Suppose the measured temperature is f(x,y) = e-x-3y2. By substitution at time t, the measured temperature is f(x(t), y(t)) = e-(x(t3)-3(y(t))2 = e-t6-3t4. so df/dt = (-6t5-12t3)e-t6-3t4. Joint cost function: qa(pa,pb) = 10-pa1/2+pb2. qb (pa,pb) = 11+2pa-10pa. dc/dpa when pa=q and pb=2 so qa=11 and qb=9 and c=17. dc/dpa = dc/dqa x dqa/dpa + dc/dqb x dqb/dpa. 17. 6: critical points constrained: (a1 . an) is critical for (x1xn) provided df/dx (a1an) = 0. If f takes a relative extrema at (a1an) then (a1an) is critical for f. Maximize f(x) = x+x3 with constraint x2=1. solution: x = -1 or 1. Definition: (x,y) is critical for the constrained problem provided there is a value of which (x,y, Fact: if (x,y) is a relative extrema for the constraint problem, it is critical.
