# A simple optimization problem

I have been attempting this problem for the last 2 hours and 45 minutes with no success and am very frustrated since it should be an extremely easy question. Please help!!!

The question takes place over 3 time periods (0,1,2). Calculating for a discount rate of 10%, what s the optimal distribution for 1200 goods. Price of the goods are $50. Cost is (Xt^2)/20. B=1+10%=1.1 The equation that I am using is: V0=pX0-c(X0) + B(pX1-cX1) + B^2(pX2-cX2) s.t. x0+x1+x2=1200 this becomes p-c'(x0)=B(p-c'x1)-B^2(p-c'x2) 50-x0/10=55-1.1(x1)/10 - 60.5 - 1.2(x2)/10 555=x0-1.1(x1)+1.21(x2) After that, I've tried everything that I can think of. The answers are supposed to be x0=409.4, x1=400.3, x2=390.3 Please help!! Thanks everyone ## Answers and Replies Related Calculus and Beyond Homework Help News on Phys.org Ray Vickson Science Advisor Homework Helper Dearly Missed I have been attempting this problem for the last 2 hours and 45 minutes with no success and am very frustrated since it should be an extremely easy question. Please help!!! The question takes place over 3 time periods (0,1,2). Calculating for a discount rate of 10%, what s the optimal distribution for 1200 goods. Price of the goods are$50. Cost is (Xt^2)/20. B=1+10%=1.1

The equation that I am using is:
V0=pX0-c(X0) + B(pX1-cX1) + B^2(pX2-cX2) s.t. x0+x1+x2=1200
this becomes
p-c'(x0)=B(p-c'x1)-B^2(p-c'x2)
50-x0/10=55-1.1(x1)/10 - 60.5 - 1.2(x2)/10
555=x0-1.1(x1)+1.21(x2)

After that, I've tried everything that I can think of. The answers are supposed to be x0=409.4, x1=400.3, x2=390.3

Thanks everyone
Are you saying that $C(x_t)= x_t^2/20$? If so, you will have some nonlinearities in the total profit:
$$\text{profit} = p x_0 - x_0^2/20 + \beta (p x_1 - x_1^2/20) + \beta^2 (p x_2 - x_2^2/20),$$
which is to be maximized, subject to the constraints $x_0 + x_1 + x_2 = 1200$ AND $x_0, x_1, x_2 \geq 0$.

One way to solve this is to neglect the inequality constraints $x_i \geq 0$ (and hope they are satisfied anyway), then use a Lagrange multiplier method to deal with the equality constraint. Another way is to reduce it to an unconstrained problem in two variables.

When $p = 50$ and $\beta = 1.1$ the Lagrange multiplier method gives $x_0 = 390.33, x_1 = 400.30, x_2 = 409.37$, which are the opposite of what you wrote! The Maple package "NLPSolve" and the EXCEL Solver tool also get this solution. Are you sure you are not supposed to have $\beta = 1/1.1$?

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Mark44
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Gary M,
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