SOLUTION: The revenue function for a particular product is R(x)= x(4-0.0001x).How to Find the largest possible revenue. to solve this do I just have to find the critical point and check i

Algebra ->  Equations -> SOLUTION: The revenue function for a particular product is R(x)= x(4-0.0001x).How to Find the largest possible revenue. to solve this do I just have to find the critical point and check i      Log On


   

Question 1205103: The revenue function for a particular product is R(x)= x(4-0.0001x).How to Find the largest possible revenue.
to solve this do I just have to find the critical point and check if its the maximum point using the derivative?
Found 3 solutions by ikleyn, MathLover1, math_tutor2020:
Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
.
The revenue function for a particular product is R(x)= x(4-0.0001x).How to Find the largest possible revenue.
to solve this do I just have to find the critical point and check if its the maximum point using the derivative?
~~~~~~~~~~~~~~~~~~~~~~~ 

This way of finding critical point of the derivative is one of possible ways to solve
the problem, but it assumes that the student knows basic of Calculus.


There are other ways that require knowledge of basics of Algebra, ONLY.


This function  R(x) = x*(4-0.0001x) is a quadratic function, which has x-intercepts at
x= 0 and x= 4%2F0.0001 = 40000.  Hence, it has the maximum half-way between the x-intercepts,
which (half-way) is  %280%2B40000%29%2F2 = 40000%2F2 = 20000.


Now, to find the largest possible revenue, simply substitute x= 20000 into the formula for R(x)
and get

    R%28x%29%5Bmax_%5D = 20000*(4-0.0001*20000) = 40000.


ANSWER.  R%28x%29%5Bmax_%5D = 40000.

Solved.

-----------------

There are other simple Algebra ways to solve this problem (and million other similar problems).

See the lessons
    - HOW TO complete the square to find the minimum/maximum of a quadratic function
    - Briefly on finding the minimum/maximum of a quadratic function
    - HOW TO complete the square to find the vertex of a parabola
    - Briefly on finding the vertex of a parabola
in this site.

Consider these lessons as your textbook,  handbook,  tutorials and  (free of charge)  home teacher.

Learn the subject from there once and for all.



Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

R%28x%29=+x%284-0.0001x%29
R%28x%29=+4x-0.0001x%5E2
using calculus:
first derivate
%28d%2Fdx%29%284x-0.0001x%5E2%29=4-2%2A0.0001x=4+-+0.0002x
equal it to zero
4+-+0.0002x=0
4+=0.0002x
x=4%2F0.0002
x=20000
plug it in given equation
R%28x%29=+4%2A%2820000%29-0.0001%2A%2820000%29%5E2=40000
max revenue is 40000

or, solve it using algebra:
R%28x%29=+4x-0.0001x%5E2
The revenue equation is a quadratic, so its graph is a parabola. Since the coefficient of the x%5E2 term is negative (-0.0001), it's an inverted parabola with the vertex at the top.
The vertex will thus be the maximum revenue.
To find the vertex, convert R to the vertex form.
R+=+a%28x-h%29%5E2%2Bk where (h,k) is the location of the vertex.
Convert to the vertex form by completing the square:
R%28x%29=+4x-0.0001x%5E2
R%28x%29=-0.0001%28+4x%2F-0.0001%2Bx%5E2%29
R%28x%29=-0.0001%28+-40000x%2Bx%5E2%29
R%28x%29=-0.0001%28+b%5E2-40000x%2Bx%5E2%29-%28-0.0001b%5E2%29...b=40000%2F2=20000
rearrange the terms in parenthesis
R%28x%29=-0.0001%28+x%5E2-40000x%2B20000%5E2%29-%28-0.0001%2A20000%5E2%29
R%28x%29=-0.0001%28+x-20000%29%5E2-%28-40000%29
R%28x%29=-0.0001%28+x-20000%29%5E2%2B40000
h=20000
k=40000
vertex is at (20000,40000)=>the maximum revenue is 40000


Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Yes that is correct.

R(x) = x(4-0.0001x)
R(x) = 4x-0.0001x^2
R'(x) = d/dx[ 4x-0.0001x^2 ]
R'(x) = 4-2*0.0001x
R'(x) = 4-0.0002x

R'(x) = 0
4-0.0002x = 0
4 = 0.0002x
x = 4/0.0002
x = 20,000

We need to check if this is a local extrema, or if it's a saddle point.
I'll use the 2nd derivative test.

R''(x) = d/dx[ R'(x) ]
R''(x) = d/dx[ 4-0.0002x ]
R''(x) = -0.0002
R''(20,000) = -0.0002

R''(x) < 0 no matter what the x input is, so R(x) is concave down throughout the entire domain.
We have determined this function has a local max.
It happens when x = 20,000.

-----------------------------

Or we could use the 1st derivative test.
Skip to the next section if you prefer the 2nd derivative test.

Draw a number line. Mark the point at x = 20,000.

Select a value to the left of that marker to plug into the 1st derivative.
Let's pick x = 0.
R'(x) = 4-0.0002x
R'(0) = 4-0.0002*0
R'(0) = 4
The actual result itself doesn't matter.
All we need to do is record whether it's positive or negative.

Now let's determine the sign for an x value larger than 20,000.
I'll pick x = 30,000.
R'(x) = 4-0.0002x
R'(30000) = 4-0.0002*30000
R'(30000) = -2

R'(x) changes from positive to negative when going from the left side of the critical value to the right side of it. As such, the tangent slopes go from positive to negative.

Draw a quick little sketch (either in your mind or do so on paper) to notice we have an upside-down parabola shape. In other words, the parabola opens downward. No graphing calculator required and there's no need to involve tables.

This downward opening parabola must mean the R(x) function has a local max. This local max is also the absolute max.

------------------------------

After we determine the max happens when x = 20000, plug this back into the original R(x) function to find the max revenue.

R(x) = x(4-0.0001x)
R(20000) = 20000(4-0.0001*20000)
R(20000) = 40,000


The largest revenue is $40,000 and occurs when selling 20,000 units. I wasn't sure what currency you are using so I went with dollars. Feel free to change to whatever other currency you're actually using.