If a cost of a function is C(x) = (49 ln x) / (28-3x), find the marginal cost when x = 7

marginal cost is

dC/dx = 49(-3x+3lnx+28)/(28-3x)^2
so at x=7, dC/dx = (7 + 3 ln7)/49

a ship sails 6 kg on a bearing 160• and then a further 10 kg on a bearing 240 as shown how far this ship from starting this point

To find the marginal cost of a function, you need to find its derivative with respect to x. Let's first find the derivative of the cost function C(x) = (49 ln x) / (28-3x).

1. Start by applying the quotient rule to differentiate the function. The quotient rule states that if you have a function f(x) = g(x)/h(x), then its derivative is given by:

f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2

Applying this rule to our cost function:

C'(x) = [((49 * 1/x) * (28 - 3x)) - (49 ln(x) * (-3))] / (28 - 3x)^2

= [(49 / x) * (28 - 3x) + 147 ln(x)] / (28 - 3x)^2

2. Now, we need to evaluate the derivative at x = 7 to find the marginal cost.

C'(7) = [(49 / 7) * (28 - 3*7) + 147 ln(7)] / (28 - 3*7)^2

= [(49 / 7) * (28 - 21) + 147 ln(7)] / (28 - 21)^2

= [(7 * 7) + 147 ln(7)] / 49

= 49 + 147 ln(7) / 49

= 1 + 3 ln(7)

Therefore, the marginal cost when x = 7 is 1 + 3 ln(7).