Cost,Revenue,and Profit: Estimate the number of scanners that would produce a maximum profit when p is the price per scanner and x is the number of scanners. p= 100-0.0001x and the cost for producing x scanners is C= 350,000+30x and the profit for producing and selling x scanners is P=R-C=xp-C. SOmehow I got P=-0.0001x^2+70x-350000. Can you verify that? I tried graphing it to get the number of scanners for the maximum profit, but my calculator doesn't work. Is there another way to find the maximum number of scanners?
To verify if the equation P=-0.0001x^2+70x-350000 is correct, we can compare it to the given information.
Given:
p = 100 - 0.0001x (Price per scanner)
C = 350,000 + 30x (Cost for producing x scanners)
P = R - C = xp - C (Profit for producing and selling x scanners)
Let's calculate P using the given expressions for p and C:
P = xp - C
P = x(100 - 0.0001x) - (350,000 + 30x)
P = 100x - 0.0001x^2 - 350,000 - 30x
Simplifying further:
P = -0.0001x^2 + 100x - 30x - 350,000
P = -0.0001x^2 + 70x - 350,000
Therefore, your equation P = -0.0001x^2 + 70x - 350,000 is correct.
Now, let's find the maximum number of scanners that would produce maximum profit. Since your calculator is not working, we can use a different method known as completing the square or find the vertex form.
The equation for profit is P = -0.0001x^2 + 70x - 350,000.
To find the maximum, we need to find the x-coordinate of the vertex of this quadratic function. The x-coordinate of the vertex can be found using the formula x = -b/2a, where a, b, and c are the coefficients of the quadratic equation.
In this case:
a = -0.0001, b = 70, and c = -350,000.
x = -b/2a
x = -70 / (2 * -0.0001)
x = -70 / -0.0002
x = 350,000
The x-coordinate of the vertex is 350,000.
Therefore, the estimated number of scanners that would produce maximum profit is 350,000 scanners.