SOLUTION: The price and the quantity x sold of a certain product obey the demand question. x=-5p+100, 0 is less than or equal to p and p is less than or equal to 20 a) express t

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Question 89589: The price and the quantity x sold of a certain product obey the demand question.

x=-5p+100, 0 is less than or equal to p and p is
less than or equal to 20
a) express the Revenue R as a function of x.
b) What is the revenue if 15 units are sold?

17) A wire 10 meters long is to be cut into two pieces. One piece will be shaped as a square and other piece will be shaped as a circle:
a) express the total area A enclosed by the pieces of wire as a function of the length x of a side of the square.
b) What is the domain of A?

Thank you sooo much!!
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
The price and the quantity x sold of a certain product obey the demand question.
x=-5p+100, 0<=p<=20
Solve for P which is price:
p=(100-x)/5
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a) express the Revenue R as a function of x.
Revenue = (price)*(quantity)
Revenue = [(100-x)/5]*x = (x/5)(100-x) = 20x-(1/5)(x^2)
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b) What is the revenue if 15 units are sold?
R(15) = 20*15 - (1/5)(15^2)
R(15) = 300 - 45 = $255
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17) A wire 10 meters long is to be cut into two pieces. One piece will be shaped as a square and other piece will be shaped as a circle:
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Let one of the pieces have length x; the other piece will have length 10-x
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a) express the total area A enclosed by the pieces of wire as a function of the length x of a side of the square.
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Circle area using the (10-x) piece of wire:
Perimeter = 10-x = 2(pi)r
r = (10-x)/2pi =
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Area of circle = (pi)r^2
Area of circle = (pi)[(10-x)/2pi]^2 = (10-x)^2/4(pi)
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Area of the square using the "x" piece of wire:
One side = (1/4)x
Area of square = side^2 = [x^2/16]
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Total Area = area of square + area of circle.
A = (x^2/16) + [(10-x)^2/4(pi)]
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b) What is the domain of A?
Domain: 0<=x<=10
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Cheers,
Stan H.