# Thread: Calculus question (app. of differentiation)

1. ## Calculus question (app. of differentiation)

Hi Guys:

1. A company produces two products whose production levels are represented by
x

and
y. As a function of these two variables profit is given by:

P(x,y)
= 900x 30x^2 + 30xy 100y 10y^2 + 2,000
a) What values of
x and y maximize the profit?
b) What is the maximum value of the profit?

c) Verify that the solution is indeed a local maximum.

Just want to make sure I am correct...

a) x = 5 and y = -25
b) -$1,750 c) I am not sure how to go about getting this 2. A local inventor has developed a new camera technology that uses special tubes to improve the quality of the pictures it takes. To produce, each camera will cost$1,000
of fixed costs, $25 for every tube used, and$400 divided by the number of tubes used
in each camera. The camera sells for $2,025. a) Write the function P(x) for the profit on a camera, where x represents the number of tubes used in each camera. b) What obvious restriction must be placed on x? c) Use the function in a) to determine the number of tubes that will maximize profit. d) Show that P(x) is maximized at the stationery point. Answers (again, tell me if and where I went wrong) a) R(x) = 2,025x C(x) = 25x + 400/x Revenue - Cost = Profit 2,000 x + 400 / x b) x > 0 I can't figure out C and D to see as I am pretty sure my equation is wrong haha. Thanks Ibrox 2. Hello, ibrox! Sorry ... your answers are way WAY off. 1. A company produces two products whose production levels are$\displaystyle x$and$\displaystyle y.$The profit is given by: .$\displaystyle P(x,y) \:=\: 900x - 30x^2 + 30xy - 100y -10y^2 + 2000$(a) What values of$\displaystyle x$and$\displaystyle y$maximize the profit? (b) What is the maximum value of the profit? (c) Verify that the solution is indeed a local maximum. (a) Set the partial derivatives equal to zero, and solve. . .$\displaystyle \begin{array}{ccccccc}P_x \:=\:900 - 60x + 30y \:=\:0 & \Longrightarrow & 4x - 2y \:=\:60 & {\color{blue}[1]} \\
P_y \:=\:30x - 100 - 20y \:=\:0 & \Longrightarrow & 3x-2y \:=\:10 & {\color{blue}[2]} \end{array}$Subtract [2] from [1]: .$\displaystyle \boxed{x \:=\:50}$Substitute into [1]: .$\displaystyle 4(50) - 2y \:=\:60 \quad\Rightarrow\quad \boxed{y \:=\:70}$To maximize profit, the company should produce: .$\displaystyle \begin{array}{c}\text{50 units of product X}\\ \text{70 units of product Y}\end{array}$(b) The maximum profit will be: . .$\displaystyle P(50,70) \:=\:900(50) - 30(50^2) + 30(50)(70) - 100(70) - 10(70^2) + 2000 \;=\;\boxed{21,\!000}$(c) To verify an extreme value, we use the Second Partials Test. Form the expression: .$\displaystyle D \:=\:(P_{xx})(P_{yy}) - (P_{xy})^2$. . and evaluate it at the critical point. If$\displaystyle D < 0$, saddle point. If$\displaystyle D > 0$, there is an extreme point. . . If$\displaystyle f_{x\!x},f_{yy} < 0$, maximum point. . . If$\displaystyle f_{x\!x},f_{yy} > 0$, minimum point. We have:. .$\displaystyle P_{xx}=\text{-}60,\;P_{yy} = \text{-}20,\;P_{xy} = 30$Hence: .$\displaystyle D \:=\:(\text{-}60)(\text{-}20) - 30^2 \:=\:+300 $. . . There is an extreme point. Since$\displaystyle P_{xx} = \text{-}60\$ (negative), it is a maximum.

3. Soroboth,

Thanks, I see what I did wrong, I screwed up getting the derivative, forgot to multiple by 2 lol.

In regards to question 2, I took a stab at it and got:

a) -25x + 400/x + 1025
b) x > 0
c) x = 4
d) Not sure what is meant by show it is maximized at stationary point.