profit queation

Question

Safe shades produce two types of sunglasses; one kind sells for $17 while the other sells for $21. The total revenue in thousands of dollars for the sale of x thousand glasses at $21 each and y thousand glasses at $17 each is given by R(x,y)= 17x+21y. The company has determined that the cost equation for these sunglasses is given by c(x,y)=4x^2-4xy+2y^2-11x+25y-3

（a) Determine the profit function.

(b) use calculus to determine how many of each type of sunglasses must be produced and sold in order to maximize profit?

(c) find the maximum profit

Answer 1

So,

px = 21

py = 17

R(x,y)=17x+21y

c(x,y)=4x^2-4xy+2y^2-11x+25y-3

a) In economics and business, Profit is the money you have made from your revenue that you do not need to return to your business via cost. In other terms, Profit = Revenue - Cost. In order to get the profit function, π(x,y), we just subtract R(x,y)-c(x,y)

π(x,y)=R(x,y)-c(x,y)=(17x+21y)-(4x^2-4xy+2y^2-11x+25y-3)

π(x,y)=-4x^2 + 4xy + 28x - 2y^2 -4y + 3

b) We must find the partial derivatives of the profit function and the critical points, then find which pair of critical points maximize the function π(x,y)

First, take the partial derivatives

π_x(x,y) = 28 + 4y - 8x

Set this to 0

28 + 4y - 8x = 0 → 28 + 4y = 8x → (28 + 4y)/8 = x*

π_y(x,y) = 4x - 4y - 4

Set this to 0

4x - 4y - 4 = 0 → 4x - 4 = 4y → (4x - 4)/4 = y* = x - 1

Check the second-order conditions to ensure that these critical points are indeed maximums

The second-order conditions are such that the second partial derivatives must be negative, and the multiple of the two second partial derivatives must be greater than the cross partial derivative squared

So, π_xx(x,y)= -8

π_yy(x,y) = -4

π_xy = π_yx = 4

(π_xx)(π_yy)>(π_yx)^2 → (-8)(-4)>(4)^2 → 32>16, confirming that this is indeed the maximum.

So, what we have just found here are the profit maximizing quantities of x and y. Further, we have two equations and two unknowns. We don't necessarily even need to use x* and y* here. Instead, we can use linear algebra and solve the system of equations

28 + 4y - 8x = 0

-4 - 4y + 4x = 0

Add R1 to R2

24 - 4x = 0 → 24 = 4x → 24/4 = x = 6

Plug this back into R1

28 + 4y - 8(6)=0 → 28 + 4y - 48 = 0 → -20 = -4y → 20/4 = y = 5

So, the profit maximizing quantities of each good is (x,y) = (6,5).

c) To find the maximized profit, we just plug these quantities into the profit function

π(x,y)=-4(6)^2 + 4(6)(5) + 28(6) - 2(5)^2 -4(5) + 3 = 77, which is the maximized profit.

I tried to go into as much detail as you could possibly need so that everything is clear, I hope this is able to help! Good luck!