
Ulkar A. answered 07/03/15
Tutor
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Engineer (Math and Electrical Engineering tutoring)
If cost equation is given as a function of x -> C(x) it follows that:
a) C(0) is the cost when x=0 -> C(0)=26*0+500-500
b)Similarly, C(10) is the cost when x=10 -> C(10)=26*10+500=760
c)p(5) is the price equation when x=5 -> p(x)=32 for x greater than or equal to 0, thus, p(5)=32
d)Assuming R(x) is revenue equation which is given by quantity*price, thus, R(x)=x*p(x)=32x
e)Assuming P(x) is the profit equation which is given by P(x)=R(x)-C(x), we can say that P(x)=32x-26x-500=6x-500