Science:Math Exam Resources/Courses/MATH105/April 2011/Question 06/Solution 1

We already have given the cost function

$C(x,y)={\frac {1}{200}}x^{2}+6x+4y+4000$

Since we sell $x$ T-Shirts of type A for $15 each and $y$ T-Shirts of type B for $10 each, the revenue must be given by

$\displaystyle R(x,y)=15x+10y$

The profit is by definition given by the formula

${\begin{aligned}P(x,y)&=R(x,y)-C(x,y)\\&=15x+10y-({\frac {1}{200}}x^{2}+6x+4y+4000)\\&=-{\frac {1}{200}}x^{2}+9x+6y-4000\end{aligned}}$

There are essentially two ways of solving the problem. Let us describe both in full detail.

Solution via Lagrange Multiplier. The fact that the company only produces 1000 T-Shirts is reflected in the constraint equation $x+y=1000$. Thus, the constraint function is given by

$\displaystyle {}g(x,y)=x+y-1000$

and the constraint translates into $g(x,y)=0$. The objective function is the profit function $P(x,y)$. We have

${\begin{aligned}P_{x}&=-{\frac {1}{100}}x+9,\\P_{y}&=6\end{aligned}}$

as well as

$\displaystyle {}g_{x}=g_{y}=1.$

Thus, for the Lagrange equation in $y$,

${\begin{aligned}P_{y}&=\lambda \cdot g_{y}\\6&=\lambda \end{aligned}}$

and for the Lagrange equation in $x$,

${\begin{aligned}P_{x}&=\lambda \cdot g_{x}\\-{\frac {1}{100}}x+9&=\lambda =6.\end{aligned}}$

We can solve these Lagrange equations to get $x=300$. Since we also must have $g(x,y)=x+y-1000=0$ we get $y=700$ as the unique solution of the Lagrange multiplier problem. The profit we make is $P(300,700)=2450$. That this is really a maximum and not a minimum can be shown be computing the profit of another distribution of T-Shirts. For instance $x=0$ and $y=1000$ yields $P(0,1000)=2000$. This argument works because we only have one global extremum under the constraint $g(x,y)=0$.

Solution without Lagrange Multipliers. The question does not state that we have to use Lagrange Multipliers to solve this problem. We can use the following solution as well. First, we solve the constraint equation $x+y=1000$ for either $x$ or $y$. Since the profit function is quadratic in $x$ and only linear in $y$ we should solve the constraint equation for $y$ to make life simpler. We get $y=1000-x$. We replace $y$ by this expression in the profit function. This results in

${\begin{aligned}P&=-{\frac {1}{200}}x^{2}+9x+6y-4000\\&=-{\frac {1}{200}}x^{2}+9x+6(1000-x)-4000\\&=-{\frac {1}{200}}x^{2}+9x+6000-6x-4000\\&=-{\frac {1}{200}}x^{2}+3x+2000\end{aligned}}$

Now the profit is transformed into a function in just $x$ and we can use methods of differential calculus learned in MATH 104 or 184 to find the global maximum. First we see that $P$ describes a parabola. Since the sign in front of the quadratic term is negative, the focal point of the parabola must be the maximum which is the only critical point of $P$. Thus, it is enough to find the zeros of $P'(x)$. We have $P'(x)=-1/100\cdot x+3$ and this equals to zero only for $x=300$. Since $y=1000-x$ we must have $y=700$.

Important Remark. If the question does not clearly mention a specific method you have to apply to solve the problem, it is your choice which method you use. Some students find the Lagrange Multiplier solution simpler, some not. Decide based on your strengths!