SOLUTION: How would I go about sketching a graph for the total cost function, revenue function, and profit function: C(x)=0.000002x^3-0.03x^2+400x+80,000 R(x)=-0.05x^2+600x p(x)=-

Algebra ->  Rational-functions -> SOLUTION: How would I go about sketching a graph for the total cost function, revenue function, and profit function: C(x)=0.000002x^3-0.03x^2+400x+80,000 R(x)=-0.05x^2+600x p(x)=-      Log On


   



Question 1195006: How would I go about sketching a graph for the total cost function, revenue function, and profit function:

C(x)=0.000002x^3-0.03x^2+400x+80,000
R(x)=-0.05x^2+600x
p(x)=-0.000002x^3-0.02x^2+200x-80,000

Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52793) About Me  (Show Source):
You can put this solution on YOUR website!
.

There are many ways to do it.

For example, a traditional way people use just for hundreds years is to sketch the polynomial curves point by point,
using tables of values, graph paper and pen / or pencil.


Another, more modern way, is to go to web-sites which have plotting tools and use them.

For example, go to web-site www.desmos.com/calculator and use its free of charge plotting tool.

All you need is to PRINT the formulas of your functions there, and you will get the curves
plotted in the next second.


Try it . . .

Regarding the revenue function R(x) = -0.05x^2+600x,  from the formula you may conclude

that it is a downward parabola with the x-interception points x= 0 and x= 600%2F0.05 = 12000

and the symmetry line  x= 6000  half-way between the x-interceptions.



The maximum of this function is at x= 6000, and you can calculate the maximum value (y-coordinate of the vertex)

by substituting x= 6000 into the formula for R(x).




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

I also recommend using technology like Desmos.
GeoGebra is another useful tool you can use.

If your teacher wants you to "plot by hand", so to speak, then you'll need to plug in various x values to find corresponding y values.
This way you'll get the (x,y) points that make up the curve.

For example, let's say we plugged in x = 10 into the C(x) function
C(x)=0.000002x^3-0.03x^2+400x+80,000
C(10)=0.000002(10)^3-0.03(10)^2+400(10)+80,000
C(10)=83997.002

You could compute that by pencil/paper, but I recommend using a calculator to do the third step.

From here we can say that the point (10, 83997.002) is on the C(x) cubic curve.
Repeat this process for other x values to generate as many points as needed.
Then draw a curve through them all the best you can.

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With regard to the revenue function R(x), we can factor out the GCF like so
R(x)=-0.05x^2+600x
R(x)=-0.05x(x-12000)

Solving R(x) = 0 has us get:
-0.05x(x-12000) = 0
-0.05x = 0 or x-12000 = 0
x = 0 or x = 12000
which are the two x intercepts.
This is where the graph crosses the x axis.

Let's rewrite the revenue equation into vertex form
Compare y = -0.05x^2+600x to y = ax^2+bx+c
a = -0.05
b = 600
c = 0

The x coordinate of the vertex is
h = -b/(2a)
h = -600/(2(-0.05))
h = 6000
Or you could remark that the x coordinate of the vertex is the midpoint of the x intercepts
h = (a+b)/2 = (0+12000)/2 = 6000
This midpoint property is directly due to symmetry.

Then plug this into the original R(x) equation to find the y coordinate of the vertex
y = -0.05x^2+600x
y = -0.05(6000)^2+600(6000)
y = 1,800,000

The vertex is located at (x,y) = (6000, 1800000)

The parabola opens downward because a < 0
Think of "negative 'a' value makes a negative frown"
Because the parabola opens downward, we know that the R(x) function maxes out at the vertex.
This helps determine the max revenue which is what many managers solely care about.

The x intercepts and vertex should give a rough idea of how the parabola looks (in a general sense).
Plug in various other x values to get more points to plot on the parabola.

Unfortunately the cubic curves aren't as easy to determine their traits about them.
Often with many cubics, it's very difficult to determine the x intercepts by hand. This is why a graphing calculator is preferred.
You could use calculus methods, but this might be beyond the scope of the course.