8. 8. 8. The weekly demand for the Pulsar 25-in. color console television is giv
ID: 2840220 • Letter: 8
Question
8. 8. 8. The weekly demand for the Pulsar 25-in. color console television is given by the demand equation p = -0.03x + 657 (0 x 12,000) where p denotes the wholesale unit price in dollars and x denotes the quantity demanded. The weekly total cost function associated with manufacturing these sets is given by C(x) = 0.000003x3 - 0.03x2 + 400x + 80,000 where C(x) denotes the total cost incurred in producing x sets. Find the level of production that will yield a maximum profit for the manufacturer. Hint: Use the quadratic formula. (Round your answer to the nearest whole number.) If exactly 206 people sign up for a charter flight, Leisure World Travel Agency charges $292/person. However, if more than 206 people sign up for the flight (assume this is the case), then each fare is reduced by $1 for each additional person. Hint: Let x denote the number of passengers above 206. Find the revenue function R(x). Determine how many passengers will result in a maximum revenue for the travel agency. What is the maximum revenue? What would be the fare per passenger in this case? The owner of the Rancho Grande has 2964 yd of fencing with which to enclose a rectangular piece of grazing land situated along the straight portion of a river. If fencing is not required along the river, what are the dimensions of the largest area he can enclose? What is this area?Explanation / Answer
The weekly demand for the Pulsar 25-in. color console television is given by the demand equation:
p = -0.03x+502
where p denotes the wholesale unit price in dollars and x denotes the quantity demanded. The weekly total cost function associated with manufacturing these sets is given by:
C(x) = 0.000001x^3-0.03x^2+400x+80,000
Revenue = xp
Profit = Revenue - Cost
Profit = x(-0.03x+502) - (0.000001x^3-0.03x^2+400x+80,000)
Profit = -0.03x^2+502x -0.000001x^3+0.03x^2-400x-80,000
differentiate the profit with respect to x
dP/dx = -0.06x +502 -0.000003x^2+0.06x -400 = 0
-0.000003x^2=-102
x^2 = 102/0.000003 = 34 000 000
x = 5830.95 = 5831 sets must be produced to maximize the profit.
d^2P/dx^2 = -0.06 -0.000006x+0.06 = -0.000006x < 0 when x=5831 which ensures that the profit has been maximized.
(2)
If exactly 206 people sign up for a charter flight, Leisure World Travel Agency charges $294/person. However, if more than 206 people sign up for the flight (assume this is the case), then each fare is reduced by $1 for each additional person. Hint: Let x denote the number of passengers above 206.
Find the revenue function R(x).
R(x) = 206(294) + 88x - x^2
To find the maximum, take the derivative and set it to 0:
R'(x) = 88 - 2x
0 = 88 - 2x
2x = 88
x = 44
So the number of passengers that denote a maximum revenue is 206+44 = 250 passengers
The revenue would be:
R(44) = (206+44)(294-44) = 250^2 = $62,500
The fare per passenger would be $250
To verify this is the maximum, let's calculate the revenue for 1 less passenger and 1 more passenger:
R(43) = (206+43)(294-43) = 249(251) = $62,499 ($1 less!)
R(45) = (206+45)(294-45) = 251(249) = $62,499 ($1 less again!)
So R(44) = 250 passengers = maximum revenue
(3)The owner of the Rancho Grande has 3036 yd of fencing with which to enclose a rectangular piece of grazing land situated along the straight portion of a river. If fencing is not required along the river, what are the dimensions of the largest area he can enclose?
The answer is a width (perpendicular to the river) of 759 and length (parallel to river) of 1518.
There are two equations: A = L x W and F.L. = 2W + L; where F.L. is fence length.
Change the second expression to solve for L:
L = F.L. - 2W = 3036-2W. Substitute this for L in the first expression:
A = (3036-2W)W = 3036W - 2W^2. If you differentiate this you get: dA = 3036-4W. Set this equal to 0:
3036-4W = 0; W = 3036 / 4 = 759.
The Long side, L = 1518. Check with some other dimensions for the short side, such as 758 yards and 760 yards just to make sure that you found the proper length of W for maximum Area.
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