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.)
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` .