Science:Math Exam Resources/Courses/MATH105/April 2011/Question 06
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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 shirts of Type A and shirts of Type B is
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.
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We already have given the cost function
Since we sell T-Shirts of type A for $15 each and T-Shirts of type B for $10 each, the revenue must be given by
The profit is by definition given by the formula
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 . Thus, the constraint function is given by
and the constraint translates into . The objective function is the profit function . We have
as well as
Thus, for the Lagrange equation in ,
and for the Lagrange equation in ,
We can solve these Lagrange equations to get . Since we also must have we get as the unique solution of the Lagrange multiplier problem. The profit we make is . 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 and yields . This argument works because we only have one global extremum under the constraint .
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 for either or . Since the profit function is quadratic in and only linear in we should solve the constraint equation for to make life simpler. We get . We replace by this expression in the profit function. This results in
Now the profit is transformed into a function in just and we can use methods of differential calculus learned in MATH 104 or 184 to find the global maximum. First we see that 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 . Thus, it is enough to find the zeros of . We have and this equals to zero only for . Since we must have .
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!