rite - cut riding lawnmowers obey the demand equation p = - 1/20x + 980. the cost of producing x lawnmowers…

rite - cut riding lawnmowers obey the demand equation p = - 1/20x + 980. the cost of producing x lawnmowers is given by the function c(x)=80x + 7000. a) express the revenue r as a function of x. b) express the profit p as a function of x. c) find the value of x that maximizes profit. what is the maximum profit? d) what price should be charged in order to maximize profit? a) r(x)= (simplify your answer. do not factor.) b) p(x)= (simplify your answer. do not factor.) c) what quantity will maximize the profit? lawnmowers what is the maximum profit? $ (round to the nearest dollar as needed.) d) what price should be charged for the maximum profit? $ (round to the nearest cent as needed.)

rite - cut riding lawnmowers obey the demand equation p = - 1/20x + 980. the cost of producing x lawnmowers is given by the function c(x)=80x + 7000. a) express the revenue r as a function of x. b) express the profit p as a function of x. c) find the value of x that maximizes profit. what is the maximum profit? d) what price should be charged in order to maximize profit? a) r(x)= (simplify your answer. do not factor.) b) p(x)= (simplify your answer. do not factor.) c) what quantity will maximize the profit? lawnmowers what is the maximum profit? $ (round to the nearest dollar as needed.) d) what price should be charged for the maximum profit? $ (round to the nearest cent as needed.)

Answer

Explanation:

Step1: Recall revenue formula

Revenue $R(x)=p\times x$. Given $p =-\frac{1}{20}x + 980$, then $R(x)=x\left(-\frac{1}{20}x + 980\right)=-\frac{1}{20}x^{2}+980x$.

Step2: Recall profit formula

Profit $P(x)=R(x)-C(x)$. Since $R(x)=-\frac{1}{20}x^{2}+980x$ and $C(x)=80x + 7000$, then $P(x)=-\frac{1}{20}x^{2}+980x-(80x + 7000)=-\frac{1}{20}x^{2}+900x - 7000$.

Step3: Find x - value for maximum profit

For a quadratic function $y = ax^{2}+bx + c$ ($a=-\frac{1}{20}$, $b = 900$, $c=-7000$), the x - value of the vertex is $x=-\frac{b}{2a}$. So $x=-\frac{900}{2\times\left(-\frac{1}{20}\right)}=9000$.

Step4: Find maximum profit

Substitute $x = 9000$ into $P(x)$: $P(9000)=-\frac{1}{20}(9000)^{2}+900\times9000 - 7000=-\frac{1}{20}\times81000000+8100000 - 7000=-4050000+8100000 - 7000 = 4043000$.

Step5: Find price for maximum profit

Substitute $x = 9000$ into the price - demand equation $p=-\frac{1}{20}x + 980$. So $p=-\frac{1}{20}\times9000+980=-450 + 980=530$.

Answer:

a) $R(x)=-\frac{1}{20}x^{2}+980x$ b) $P(x)=-\frac{1}{20}x^{2}+900x - 7000$ c) 9000 lawnmowers, $4043000$ d) $530.00$