Question

find the production level that will maximize profit.If  C(x) = 15000 + 600x − 2.8^2+ 0.004x^3 is the cost function and  p(x) = 4200 − 7x  is the demand function, find the production level that will maximize profit. (Hint: If the profit is maximized, then the marginal revenue equals the marginal cost.)

Accepted Answer

You need to remember that the marginal cost equals marginal revenue means that C'(x) = p'(x), hence you should find the derivative of C(x) and p(x) with respect to x such that:
C'(x) = 600 - 5.6x + 0.012x^2
p'(x) = -7
You need to set the equations of C'(x) and p'(x) equal such that:
600 - 5.6x + 0.012x^2 = -7
0.012x^2 - 5.6x + 607 = 0
Using quadratic formula yields:
x_(1,2) = (5.6+-sqrt(31.36 - 29.136))/0.024
x_(1,2) = (5.6+-1.4)/0.024
x_1 = 7/0.024 =gt x_1 = 291.6
x_2 = 175
Hence, evaluating the values of x for the marginal cost equals marginal revenue yields x_1~~ 292  and x_2 = 175 .

Answer

The cost function is C(x)=15000+600x-2.8x^2+0.004x^3  and the revenue function is P(x)=4200-7x
Here we have to find the production level that will maximize the profit.
For maximum profit, the marginal cost should be equal to the marginal revenue.
Now, the marginal cost is the derivative of cost function.
C'(x)=600-5.6x+0.012x^2
Similarly, the marginal revenue is the derivative of the revenue function.
P'(x)=-7
When we equate both, we get,
C'(x)=P'(x)
600-5.6x+0.012x^2=-7
0.012x^2-5.6x+607=0
Now,
x=\frac{5.6\pm \sqrt{5.6^2-4(0.012)(607)}}{2(0.012)}
  =\frac{5.6\pm \sqrt{2.224}}{0.024}
So x=295.46 \ or \ x=171.21
In other words, x\approx 295\ units \ or \ x=171\ units

Answer

The cost function for production of x units is given by C(x) = 15000 + 600x − 2.8^2+ 0.004x^3. The revenue earned for x units sold is given by P(x) = 4200 − 7x .
If x units of a product are being produced, marginal cost is defined as the change in revenue if x+1 units are produced. This is the derivative of the cost function or C'(x). Similarly, marginal revenue is given by P'(x).
If the profit is maximized when X units are being produced, at x=X, C'(x)=P'(x).
C'(x) = 600-5.6x+.012x^2
P'(x) = -7
Equating the two,
C'(x)=P'(x)
=> 600-5.6x+.012x^2= -7
=> .012x^2-5.6x+607=0
The quadratic equation obtained can be solved with the quadratic function.
X = (-b+-sqrt(b^2-4*a*c))/(2a)
X = (5.6+- sqrt(5.6^2-4*.012*607))/(2*0.012)
or X= 295.47 and X = 171.19.
As production has to be in whole units, profit is maximized when either 295 or 171 units are produced.

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