Science:Math Exam Resources/Courses/MATH105/April 2012/Question 06 (b)

MATH105 April 2012

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[hide] Question 06 (b)
According to market research, the demand curve for a local pizza restaurant satisﬁes the following relation: if p is the price of a pizza (in dollars), and q is the number of pizzas sold per day, then $\displaystyle p^{2}+4q^{2}=800.$ The restaurant owners want to determine what price the restaurant should charge for each pizza in order to make their daily revenue as high as possible. (b) Use the method of Lagrange multipliers to solve the problem in part (a). There is no need to justify that the solution you obtained is the absolute maximum or minimum. A solution that does not use the method of Lagrange multipliers will receive no credit, even if the answer is correct.

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[show] Hint
Recall that the objective function is $f(p,q)=pq$ , with contraint $g(p,q)=p^{2}+4q^{2}=800$ . Solve the constraint equation for zero to obtain h(p,q) and set up the Lagrange system $\nabla f(p,q)=\lambda \nabla h(p,q)$ and h(p,q) = 0. Now, compute the partial derivatives of both sides with respect to the variables, p and q to set up the system of equations that need to be solved.

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[show] 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 are maximizing $f(p,q)=pq$ subject to $g(p,q)=p^{2}+4q^{2}=800$ . Solving the constraint for zero we obtain $h(p,q)=p^{2}+4q^{2}-800=0$ , so that the Lagrange system is $\nabla f(p,q)=\lambda \nabla h(p,q)$ Computing the gradients, $f_{p}=(pq)_{p}=q\qquad f_{q}=(pq)_{q}=p$ Also, $h_{p}=(p^{2}+4q^{2})_{p}=2p\qquad h_{q}=(p^{2}+4q^{2})_{q}=8q$ Therefore, $\displaystyle <q,p>=\lambda <2p,8q>$ which gives us two equations, namely: $\displaystyle {\begin{aligned}q&=2\lambda p\quad {\text{(equation 1)}}\\p&=8\lambda q\quad {\text{(equation 2)}}\end{aligned}}$ In substituting $q=2\lambda p$ into $p=8\lambda q$ , we get $\displaystyle p=8\lambda (2\lambda p)=16\lambda ^{2}p$ We can factor this to find $p(1-16\lambda ^{2})=0$ so that either $\displaystyle p=0$ $\displaystyle 1-16\lambda ^{2}=0$ , i.e. $\lambda =\pm 1/4$ . If $p=0$ then $q=2\lambda (0)=0$ , and $p=q=0$ cannot satisfy the constraint. We reject this case. If $\lambda =\pm 1/4$ then $q=2(\pm 1/4)p=\pm p/2$ . Substituting this into the constraint gives: $0=p^{2}+4(\pm p/2)^{2}-800=2p^{2}-800$ which mean p² = 400, or $p=\pm 20$ . We take the positive root (since price should be positive). At a price of $20 (with a demand of 10/day), the revenue is maximized. Aside: it wouldn't make sense for either p or q to be 0 (there would be no revenue), so we can safely divide the equations (1) and (2) without the risk of dividing by 0. Or to look at it another way: if q = 0 then from the equations that forces p=0 (and vice versa), and p=q=0 could not satisfy the constraint. If we divide the equations then $q/p=(2\lambda p)/(8\lambda q)=p/(4q)$ . Cross-multiplying gives the relation $p^{2}=4q^{2}$ which could be used in the constraint equation to find p and q.