The telephone company is planning to introduce two new types of executive communications systems that it hopes to sell to its largest commercial customers. It is estimated that if the first type of system is priced at x hundred dollars per system and the second type at y hundred dollars per system, approximately 40−8x+5y consumers will buy the first type and 50+9x−7y will buy the second type. If the cost of manufacturing the first type is $1000 per system and the cost of manufacturing the second type is $2900 per system, what prices x and y will maximize the telephone company's profit?

x = 31 hundred dolars   and      

y = 91/2 = 45.5 hundred dolars

Step-by-step explanation:

Given

R(x) = (40−8x+5y)*x + (50+9x−7y)*y

C(x) = (40−8x+5y)*10 + (50+9x−7y)*29

We can use the equation

P(x) = R(x) - C(x)

where

P(x) is the profit

R(x) is the revenue

and C(x) is the costs

In order to maximize the telephone company's profit, we apply

P'(x) = R(x)' - C(x)' = 0

⇒ R(x)' = ((40−8x+5y)*x + (50+9x−7y)*y)' = (40x-8x²+14xy+50y-7y²)'

⇒ C(x)' = ((40−8x+5y)*10 + (50+9x−7y)*29)' = (1850+181x-153y)'

⇒ P'(x) = -8x²-7y²-141x+203y+14xy-1850

The first-order partial derivatives of these functions are

Px(x,y) = -16x-141+14y

Py(x,y) = -14y+203+14x

Setting these equal to zero and solving we obtain:

-16x+14y-141 = 0

14x-14y+203=0

we get the solution

x = 31     and       y = 91/2 = 45.5

Finally, the company should produce  3100  units of the first system, and  4550 units of the second system.

this shows a reflection over the x-axis and is translated one unit down and 6 units to the right (x + 6, y - 1)

hope this !

~just a girl in love with shawn mendes