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 $0.60 each. for each nickel increase in the price, the sales per month drop by 1000 cans.\n(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.\n(b) find a graph of r(x) that clearly shows a maximum for r(x).\n(c) how much should the market charge per can to realize the maximum revenue? what is the maximum revenue?\n\n(a) r(x) = \frac{}{1000 - 0.5x}\n(b) choose the correct graph below\na. graph image\nb. graph image\nc. graph image\nd. graph image\nall 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 $0.60 each. for each nickel increase in the price, the sales per month drop by 1000 cans.\n(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.\n(b) find a graph of r(x) that clearly shows a maximum for r(x).\n(c) how much should the market charge per can to realize the maximum revenue? what is the maximum revenue?\n\n(a) r(x) = \frac{}{1000 - 0.5x}\n(b) choose the correct graph below\na. graph image\nb. graph image\nc. graph image\nd. graph image\nall 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 is $0.60$ and the initial quantity is $23000$. The price per - can $p(x)=0.60 + 0.05x$, and the quantity of cans sold $q(x)=23000-1000x$.

Step2: Find the revenue function

Revenue $R(x)=p(x)\times q(x)=(0.60 + 0.05x)(23000 - 1000x)$. Expand the product: [ \begin{align*} R(x)&=0.60\times23000-0.60\times1000x+0.05x\times23000-0.05x\times1000x\ &=13800-600x + 1150x-50x^{2}\ &=- 50x^{2}+550x + 13800 \end{align*} ]

Step3: Find the vertex of the quadratic function

For a quadratic function $y = ax^{2}+bx + c$ ($a=-50$, $b = 550$, $c = 13800$), the $x$ - coordinate of the vertex is $x=-\frac{b}{2a}$. [ x=-\frac{550}{2\times(-50)}=\frac{550}{100}=5.5 ]

Step4: Find the maximum price and revenue

The price per can to maximize revenue is $p(5.5)=0.60+0.05\times5.5=0.60 + 0.275=$0.875$. The maximum revenue is $R(5.5)=-50\times(5.5)^{2}+550\times5.5 + 13800$. [ \begin{align*} R(5.5)&=-50\times30.25+3025+13800\ &=-1512.5+3025+13800\ &=15312.5 \end{align*} ] For part (b), since $a=-50\lt0$, the graph of $y = R(x)$ is a parabola opening down - ward.

Answer:

(a) $R(x)=-50x^{2}+550x + 13800$ (b) The graph of a parabola opening downward. Among the options, the correct one is the graph that is a parabola opening downward (you need to visually identify the correct one from the given options A, B, C, D based on the description of a downward - opening parabola). (c) The market should charge $$0.875$ per can. The maximum revenue is $$15312.5$