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

MATH105 April 2011

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Question 06

A certain T-shirt company makes two types of shirts. Type A says "Much Music rocks" while Type B says "Business Calculus rocks". The company produces 1000 T-shirts in total every year. It sells each Type A shirt at $15 and each Type B shirt at $10. If the cost of producing $x$ shirts of Type A and $y$ shirts of Type B is $C={\frac {1}{200}}x^{2}+6x+4y+4000,$ determine how many shirts of each kind the company should produce to maximize its profit every year. Give a simple justification why the answer you obtained gives the maximum and not the minimum.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
The profit $P$ is given by $P=R-C$, where $R$ denotes the revenue and $C$ the costs. Decide what the objective and constraint function could be What method do you know to solve an optimization problem in two variables?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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!

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