Asked by jan
the price p and the quantity x sold of a certain product obey the demand equation
p=(-1/3)x + 100
o is < or equal to x < or equal to 300
A) express revenue R as a function of x
b) what quantity x maximizes revenue.
p=(-1/3)x + 100
o is < or equal to x < or equal to 300
A) express revenue R as a function of x
b) what quantity x maximizes revenue.
Answers
Answered by
bobpursley
Revenue=p*x
you finish..
Take the deriviative of R with respect to x, set to zero, solve for x.
you finish..
Take the deriviative of R with respect to x, set to zero, solve for x.
Answered by
jan
so would the revenue be
(-1/3)x + 100 * 300
I don't understand- how would you multiply by x is its inequalities?
Also, what is the deriviative?
(-1/3)x + 100 * 300
I don't understand- how would you multiply by x is its inequalities?
Also, what is the deriviative?
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