Find the maximum profit and the number of units that must be...

Find the maximum profit and the number of units that must be produced and sold in order to yield the maximum profit. Assume that revenue, $R(x),$ and $\cos t, C(x)$ are in dollars.
$$R(x)=50 x-0.5 x^{2}, \quad C(x)=4 x+10$$

Key Concepts

Quadratic Function Optimization

When the profit function is quadratic, as is common in many economic scenarios, it typically takes the form ax² + bx + c. Since quadratic functions have a single maximum (or minimum) point, finding the vertex of the parabola gives the optimal solution. This is particularly useful because the vertex represents either the highest profit (if the parabola opens downward) or the lowest cost (if it opens upward).

Using Derivatives for Optimization

Calculus offers a method for finding maximum or minimum values through derivatives. By taking the derivative of the profit function with respect to the number of units produced and setting it equal to zero, one can solve for the critical points. These points are then analyzed to confirm that they correspond to a maximum profit value.

Profit Function

The profit function represents the difference between revenue and cost functions. In optimization problems, it is crucial to express profit as a function of the number of units produced and sold. Once formulated, the profit function can be analyzed to determine the level of production that maximizes profit.

Transcript

00:03 Okay, in number 23, there's some things that we know and some things that we need to find out.

00:09 I'm kind of kind of kind of try to get organized here with all our different terms, right? price per unit, total cost, total revenue, profit.

00:17 These are all defined in the book, right? and they're also defined in the problem.

00:22 Okay, so total cost is like how much money you're going to spend to create something.

00:25 Total revenue is how much money you're going to bring in to sell it.

00:28 So it makes sense that the profit is your total revenue, minus your total cost.

00:33 Right? and we're trying to find the maximum profit.

00:35 So we need to mess with this profit equation a little bit, and then we can find a minimum.

00:41 If we go ahead and use the definition of our profit function, then we can go ahead and put in the two parts, this and this for the profit equation.

00:54 So i will have 50x minus 0 .5, just so we don't lose the decimal.

01:02 X squared minus and remember it's minus this whole thing so to make sure our negatives that don't get messed up i'll do 4x plus 10 in parentheses i mean because everything is going to be flipped here all right so when i distribute that it's going to be minus 4x minus 10 so bringing my xs together i have 46x and i still have my planal x squared and then i've gotten minus 10, okay? these are in kind of a funky order, but it doesn't really matter because we're going to take the derivative of this.

01:44 Why do we need to find the derivative of p? well, this tells us this derivative.

01:52 What the slope of the line is for the graph, and at the point where it's at its highest, or it's lowest, but it's at its highest for this equation, it's going to have a slope of zero.

02:04 And the derivative is equal to the slope on this equation...

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