To produce a can of soda the production cost per day for...

To produce a can of soda, the production cost per day for Company Refreshas is given by: C(x) = 100 + 9x - 0.01x^2 + 0.005x^3. Evaluate the maximum cost the company incurs in a day. Determine the marginal cost when x = 32. Suppose Company Refreshas could produce a maximum of 15000 cans of sodas in a day, then its profit in thousands of dollars is given by: P(x) = 34512 + 603200x - 754x^2. Determine how much profit Company Refreshas makes. Calculate the number of cans of sodas to sell to maximize profit.

Transcript

00:01 The marginal cost of producing x units for one day of a certain product is the marginal cost mc equals 16x minus $1 ,591 where the cost of production is in dollars.

00:13 The selling price is fixed at nine per unit and the fixed cost is $1 ,800 per day.

00:18 We want to find the profit if 50 units are sold.

00:21 So the profit with respect to x, the profit function is the revenue function minus the cost function.

00:30 So we know the revenue function, our x, can be found by looking at the selling price which is nine per unit and the fixed is nine per unit and then we also need to when we're dealing with the cost function we have our $1 ,800 per day.

00:53 So that wouldn't have anything to do with the revenue.

00:56 Our revenue would just be nine per unit so it'd be nine and we're producing x units.

01:03 So there's our revenue.

01:05 So now our cost function is going to take a little bit more to get.

01:09 We have our marginal cost function.

01:12 The marginal cost of 16x minus $1 ,591.

01:20 The derivative of the cost function which is what we want is the marginal cost.

01:26 So the derivative of the cost equals the marginal cost.

01:31 So our derivative of our cost is 16x minus $1 ,591.

01:37 We can integrate both sides with respect to x and that would give us our cost function.

01:48 So the cost function would be the integral of 16x minus $1 ,591.

01:54 So we need to do the integral of 16x minus $1 ,591.

02:00 So the integral of 16x that would be our constant 16.

02:07 We take our x and we bring that up a degree to x squared and we have to divide by our new degree of two using the power rule for integration minus the integral of $1 ,591 and i forgot our dx here.

02:21 We're doing this with respect to x but the integral nonetheless of this constant would just be that constant times x and we have to remember our plus c our constant of integration...

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