M is not equal to m
P is not equal to p
Task (1)
The following four 4 equations are needed to be plotted as CONTOUR lines on the axes space (C, T) where C is the horizontal axis, and T is the vertical axis.
Contour lines for each equation should be between 6 and 8.
In total there should be five (5) plots, one for each equation, and the fifth which, is the most important, a plot with all the four equations are imposed on the same axes system (C, T)
M = (C) * ( 2*T + 0.028) (eq. 1)
P1 = (0.1237) * (C^(-4/3)) * (T) * ((T+1)^(1/3)) (eq. 2)
P2 = (0.003) * (C) * (T+1) (eq. 3)
P3 = 4 * (C^2) * (T) * (T+1) (eq 4)
M contour lines to be dashed lines, while P1, P2 and P3 to be solid lines in different colours.
No area filling, and no labeling are required.
Task (2)
Find the boundary between each two of the equations P1, P2, P3 by solving:
P1 = P2 , P1 = P3 , P2 = P3
Plot the boundary lines on a single figure that also includes the contours of M, P1, P2 and P3.
Task (3)
Solve the optimality condition for each P equation against M according to the following relation:
Condition for Optimality: ∇M = λ ∇P with respect to C and T.
Note that the ∇is the gradient vector, and λ is a scalar (the Langrange multiplier) so that:
M is the objective function and P is the constraint function.
The solution is obtained by differentiating M and P twice ( that is one time with respect to each variable C and T.
Then finding the values of λ. And finally substituting in the constraint function P and find the values for C and T. (( In this method we assume that ∇P is NOT equal to zero.))
Thus, solve for C and T:
dM/dC = λ (dP/dC) and dM/dT = λ (dP/dT)
M will be the objective function, and P will be the constraint function.
Note The total solution must be repeated three times:
∇M = λ ∇P1
∇M = λ ∇P2
∇M = λ ∇P3