a market sells cans of soda pop in machines. it finds that sales average 23,000 cans per month when the cans…

a market sells cans of soda pop in machines. it finds that sales average 23,000 cans per month when the cans sell for 50¢ each. for each nickel increase in the price, the sales per month drop by 100 cans. (a) determine a function r(x) that models the total revenue realized by the market, where x is the number of $0.05 increases in the price of a can. (b) find a graph of r(x) that clearly shows a maximum for r(x). (c) how much should the market charge per can to realize the maximum revenue? what is the maximum revenue? (b) choose the correct graph below. all graphs are shown in a 0,15,1 by 10000,21000,1000 viewing window.

a market sells cans of soda pop in machines. it finds that sales average 23,000 cans per month when the cans sell for 50¢ each. for each nickel increase in the price, the sales per month drop by 100 cans. (a) determine a function r(x) that models the total revenue realized by the market, where x is the number of $0.05 increases in the price of a can. (b) find a graph of r(x) that clearly shows a maximum for r(x). (c) how much should the market charge per can to realize the maximum revenue? what is the maximum revenue? (b) choose the correct graph below. all graphs are shown in a 0,15,1 by 10000,21000,1000 viewing window.

Answer

Explanation:

Step1: Determine price and quantity functions

The initial price per can is $0.50$ and the initial quantity sold is $23000$. For each $x$ (number of $0.05$ - increases in price), the price per can $p(x)=0.50 + 0.05x$ and the quantity sold $q(x)=23000-100x$.

Step2: Find the revenue function

Revenue $R(x)$ is the product of price and quantity, so $R(x)=p(x)\times q(x)=(0.50 + 0.05x)(23000 - 100x)$. Expand the product: [ \begin{align*} R(x)&=0.50\times23000-0.50\times100x+0.05x\times23000-0.05x\times100x\ &=11500 - 50x+1150x-5x^{2}\ &=- 5x^{2}+1100x + 11500 \end{align*} ]

Step3: Find the vertex of the parabola

For a quadratic function $y = ax^{2}+bx + c$ (here $a=-5$, $b = 1100$, $c = 11500$), the $x$ - coordinate of the vertex is $x=-\frac{b}{2a}$. [ x=-\frac{1100}{2\times(-5)}=\frac{-1100}{-10}=110 ]

Step4: Find the maximum - revenue price

The price per can to maximize revenue is $p(110)=0.50+0.05\times110=0.50 + 5.50=6$ dollars.

Step5: Find the maximum revenue

Substitute $x = 110$ into the revenue function $R(x)=-5x^{2}+1100x + 11500$. [ \begin{align*} R(110)&=-5\times(110)^{2}+1100\times110 + 11500\ &=-5\times12100+121000+11500\ &=-60500+121000+11500\ &=72000 \end{align*} ] For part (b), since $R(x)=-5x^{2}+1100x + 11500$ is a quadratic function with $a=-5\lt0$, the graph is a parabola opening downwards.

Answer:

(a) $R(x)=-5x^{2}+1100x + 11500$ (b) The graph of a parabola opening downwards (the one that has a maximum point in the given viewing - window). Without seeing the actual details of the graphs A, B, C, D, we know it should be a downward - opening parabola. (c) The market should charge $6$ dollars per can to realize the maximum revenue. The maximum revenue is $72000$ dollars.