Given the weekly demand curve of a local wine producer is p= 50-0.1q, and that the total cost function is c= 1500+ 10q, where q bottles are produced each day and sold at a price of $p per unit. a)Express the weekly profit as a function of price. b) at what price must a bottle of wine be sold to realise maximum profit. c) what is the maximum profit that ca be made by the producer

Step-by-step explanation:

From the given information:

a) To express the weekly profit as a function of price

Cost =C(q) = 1500 + 10q

Revenue = p×q = (50 − 0.1q)×q = 50q - 0.1q²

Revenue = 50q - 0.1q²

Weekly profit = Revenue - Cost

P(q) = (50q -0.1q²) - (1500 + 10q)

P(q)= -0.1 q² + 40 q - 1500  

However, q = 500 - 10 p using p = 50 − 0.1q

P= -0.1 (500 - 10 p)² + 40 (500 - 10 p) - 1500

P= -10 p² + 600 p - 6500

b)

The price at which the bottle of the wine must be sold to realise a maximum profit can be determined by finding the derivative and then set it to 0  

P' = 0

= -20p+600 = 0

20p = 600

p = 600/20

p = $30

c)

The maximum profit that can be made by the producer is:

P= -10(30)² + 600(30) - 6500

P = - 9000 + 18000 - 6500

P = $2500

i think it's ф or x=6,-6

step-by-step explanation:

-x^2=-36

x^2=-36

ф

or

if you mean by : -(x^2)=-36

it'll be: x^2=36

x=6,x=-6

hope it'll ! have a good day!

you for asking!

da freeck kinda math is this just drop out

step-by-step explanation: