+95 Let p(x)=75-(x^2/4) be the price in dollars per cake that a bakery can charge if they want to sell x cakes. What cake price will maximize revenue? (Recall, revenue is the total amount of money received from the sale of x cakes. Make sure to justify why you answer corresponds to an absolute maximum.)
+129907 Note.....revenue = quantity * price = x * p(x) and the revenue function, R(x) is :
R(x) = x* p(x) = x [75 - (1/4)x^2] = 75x - (1/4)x^3 take the derivative
R'(x) = 75 - (3/4)x^2 set this to 0
75 - (3/4)x^2 = 0
75 = (3/4)x^2 multiply both sides by 4/3
100 = x^2 take the positive root
10 = x so ....selling 10 cakes maximizes the revenuue
The second derivative = -(4/3)x
And subbing the critical point of x = 10 ito this produces a negative......which indicates a max
And the max revenue is shown by this graph : https://www.desmos.com/calculator/i1wj8zvt4k
The max revenue = $500
So.....the cake price which produces the max revenue is : 75 - (1/4)(10)^2 = 75 - (1/4)(100) = 75 - 25 = $50
Which makes sense....!!!
