hunting activities whose age is x. What is the age at which there are the most h
ID: 2791186 • Letter: H
Question
hunting activities whose age is x. What is the age at which there are the most hunters? Approximately how many hunters are this age? 2. The function H(x) -3.24x2 +242.1x-738.4 models the number of individuals who engage in 3. The monthly revenue R achieved by selling x wristwatches is figured to be R(x)-75x-0.2. The monthly cost C of selling x wristwatches is C(x)-32x +1750. a. What is the maximum revenue? b. Profit is given as P(x) R(x)-Cx). What is the maximum profit? What is the cost of producing wristwatches that gives maximum profit? . The cost of running a business is given by C(x)- x-6x-7 and the revenue for the same busine s R(x) 3x+3. x represents the number packages of widgets and each package contains 100 widg nd profit is in thousands of dollars. a. Find the x-value of the right most break-even point. Find the maximum profit of this business, and the number of widgets that must be ma and sold to maximized profit. b.Explanation / Answer
Break even point is where revenues equal costs
or,
x^2-6x-7=3x+3
=>x^2-9x-10=0
=>(x+1)*(x-10)=0
Hence, x=-1 or x=10
So, right most breakven point is x=10 as quantity, x cannot be negative
Profit=Revenue-Cost
Profit=3x+3-(x^2-6x-7)
Profit=9x+10-x^2
To maximize profit, differentiation prfit w.r.t. x, we get
9-2x
Setting this to zero, we get
9-2x=0
=>x=4.5
As x is packet of widges and each packet contains 100 widges, it means 4.5*100=450 widges so no decimal
To check whether this is actually maximum or not, we double differentiaite profit, we get
-2
As double differentiation is negative, we are sure that at x=4.5, profit is maximum
As calculated above at x=4.5, profit is maximum that is the firm must sell 4.5 packets of widgets or 4.5*100=450 widgets
Profit=9x+10-x^2
At x=4.5, profit is maximum so Maximum profit=9*4.5+10-4.5^2
Maximum profit=30.25
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.