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Question

I'm having trouble with this one and I need some help. Thank you.

 

In the following exercise, the revenue and cost functions are given.

R(x) = 20x - 0.1x^2

C(x) = 4x + 2

Where x is the number of staplers, and R(x) and C(x) are in dollars.

 

a. Find the profit function of P(x).

 

b. Find the number of staplers which need to be sold in order to maximize profit.

 

c. Find the maximum profit.

Answer 1

Hi Jeff,

 

We'll take each question step-by-step.

 

 

Part A (Profit Function):

 

Let's begin by expressing the Profit Function, P(x), in terms of the Revenue Function, R(x), and the Cost Function, C(x):

    P(x) = R(x) - C(x)

 

Next, let's plug in the equations for R(x) and C(x) to get P(X):

    P(x) = (20 - 0.1x^2) - (4x + 2)

    P(x) = 20 - 0.1x^2 - 4x - 2

    P(x) = -0.1x^2 + 16x - 2 (Profit Function)

 

 

Part B (# of Staples Sold for Max. Profit):

 

Let's begin by writing the standard form of a quadratic function and our Profit Function from Part A and then determine the coefficients (a,b,c) for our parabolic function:

    ax^2 + bx + c                [standard form of quadratic function]

    -0.1x^2 + 16x + c           [Profit Function from Part A]  

    a = -0.1; b = 16; c = -2

 

Since the sign of the x^2 term is negative, our parabola will open down and our vertex will be a maximum.  The x coordinate of the vertex can be calculated by plugging in our coefficient values for a and b into the axis of symmetry formula :

    x = -b/2a          [axis of symmetry formula]

    x = -16/2(-0.1)  [our a and b values plugged in]

    x = -16/-0.2

    x = 80 (# of Staples Sold for Max. Profit)

 

 

Part C (Max. Profit):

 

We can calculate the y coordinate of the parabola's vertex (max. value) by plugging into our Profit Function from Part A - the vertex x coordinate obtained in Part B:

    -0.1x^2 + 16x - 2          [Profit Function from Part A]

    -0.1(80)^2 + 16(80) - 2  [vertex x coordinate obtained in Part B]

     y = -0.1(6400) + 1280 - 2

     y = -640 + 1280 - 2

     y = 640 - 2

     y = $638 (Max. Profit)

 

 

Finally, below is the link to our graph generated on a graphing calculator showing our Profit Function:

 

https://dl.dropbox.com/s/feoo4rubd93zhsj/Graph_of_Profit_Function.png?raw=1

 

The graph shows the Profit in y dollars vs. the Sales of x staples with the coordinates of the vertex (80,638) confirming our answers to

Part B (x coordinate) and Part C (y coordinate) for the maximum value of the function.

 

Thanks for submitting this problem and glad to help.

 

God bless, Jordan.

Answer 2

a. Profit is revenue minus cost, so P(x) = 20x - 0.1x^2 - 4x - 2 = 16x - 0.1x^2 - 2

 

b. To find the maximum of a function, take the first derivative: P'(x) = 16 - 0.2x

Then, set the derivative equal to zero, and solve for x: 16 - 0.2x = 0; x = 80

So, 80 staplers need to be sold in order to maximize profit.

 

c. To find the maximum profit, plug x=80 into the profit function: P(80) = 16 * 80 - 0.1 * 80^2 - 2 = 638 dollars.