Verifying exact value of optimal solution

  • Thread starter Thread starter csc2iffy
  • Start date Start date
  • Tags Tags
    Value
csc2iffy
Messages
74
Reaction score
0

Homework Statement



(a) Formulate a LP model for the problem.

x = # units special risk insurance
y = # units mortgages
P = profit

Objective: Maximize P = 5x + 2y

Constraints:
3x + 2y ≤ 2400​
y ≤ 800​
2x ≤ 1200​
x,y ≥ 0​

(b) Use graphical method to solve.
I graphed the constraints and saw that the maximum value point within the feasible region is (600,300). Therefore, max P = 5(600)+2(300) = 3600

(c) Verify the exact solution of your optimal solution from part (b) by solving algebraically for the simultaneous solution of the relevant two equations.
Here, I am not sure what the "relevant two equations" are.

Homework Equations





The Attempt at a Solution


This is my attempt at (c):
3x + 2y = 2400 --> y = (1/2)(2400-3x)
2x = 1200 --> x = 600
y = (1/2)(2400 - 3(600)) = 300
Then P = 5x + 2y = 5(600) + 2(300) = 3600

Is this correct?
 
Physics news on Phys.org
csc2iffy said:

Homework Statement



(a) Formulate a LP model for the problem.

x = # units special risk insurance
y = # units mortgages
P = profit

Objective: Maximize P = 5x + 2y

Constraints:
3x + 2y ≤ 2400​
y ≤ 800​
2x ≤ 1200​
x,y ≥ 0​

(b) Use graphical method to solve.
I graphed the constraints and saw that the maximum value point within the feasible region is (600,300). Therefore, max P = 5(600)+2(300) = 3600

(c) Verify the exact solution of your optimal solution from part (b) by solving algebraically for the simultaneous solution of the relevant two equations.
Here, I am not sure what the "relevant two equations" are.

Homework Equations





The Attempt at a Solution


This is my attempt at (c):
3x + 2y = 2400 --> y = (1/2)(2400-3x)
2x = 1200 --> x = 600
y = (1/2)(2400 - 3(600)) = 300
Then P = 5x + 2y = 5(600) + 2(300) = 3600

Is this correct?

When you solve the problem graphically, the solution point will lie on the intersection of two of the 5 boundary lines L1 --L5, where L1 is the line {3x + 2y = 2400}, L2 is {y = 800}, L3 is {2x = 1200}, L4 is {x = 0} and L5 is {y = 0}. The two relevant intersection lines give you your two relevant equations.

RGV
 
csc2iffy said:

Homework Statement



(a) Formulate a LP model for the problem.

x = # units special risk insurance
y = # units mortgages
P = profit

Objective: Maximize P = 5x + 2y

Constraints:
3x + 2y ≤ 2400​
y ≤ 800​
2x ≤ 1200​
x,y ≥ 0​



(b) Use graphical method to solve.
I graphed the constraints and saw that the maximum value point within the feasible region is (600,300). Therefore, max P = 5(600)+2(300) = 3600

(c) Verify the exact solution of your optimal solution from part (b) by solving algebraically for the simultaneous solution of the relevant two equations.
Here, I am not sure what the "relevant two equations" are.

Homework Equations





The Attempt at a Solution


This is my attempt at (c):
3x + 2y = 2400 --> y = (1/2)(2400-3x)
2x = 1200 --> x = 600
y = (1/2)(2400 - 3(600)) = 300
Then P = 5x + 2y = 5(600) + 2(300) = 3600

Is this correct?


I think you can set all the inequalities as equalities, and solve each pairs of them and substitute into P to see which solution gives you the maximum. The idea is that this is a totally linear optimization, the maximum or minimum ALWAYS occurs at the "corner" of your feasible region
 
Once he has solved the problem graphically he can see exactly which of the pairs hold, so there is no need to try them all. Of course, the simplex method uses the "corner point" property to search intelligently through the relevant corners in general problems of this type.

RGV
 
Thanks!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top