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.
$$R(x)=5 x, \quad C(x)=0.001 x^{2}+1.2 x+60$$

Key Concepts

Revenue Function

A revenue function represents the total income from selling a certain number of units of a product. It is typically expressed as a function of the number of units sold and is fundamental in economic analysis for understanding how sales volume impacts overall earnings.

Cost Function

A cost function quantifies the total cost incurred in producing a given number of units. It often includes fixed costs, variable costs, and sometimes can be expressed as a polynomial function when variable costs increase at a non-constant rate. This function is essential for determining profitability.

Profit Function

The profit function is defined as the difference between the revenue function and the cost function. It indicates the net gain from production and sales, and its maximization is a common goal in business operations to ensure optimal use of resources.

Quadratic Optimization

Quadratic optimization involves finding the maximum or minimum of a quadratic function. When the profit function is quadratic and opens downward (i.e., the coefficient of the squared term is negative), its maximum value occurs at the vertex of the parabola. This concept is vital in many economic and business applications where optimal decision-making is required.

Differentiation for Optimization

Differentiation is a calculus technique used to find the rate at which a function changes. For optimization problems, taking the derivative of a function and setting it equal to zero helps find critical points, which can then be evaluated to determine whether they represent maximum or minimum values. This process is especially useful in profit maximization scenarios.

Transcript

00:01 Okay, on number 26, number 26, we have the following information, total cost, total revenue, and we are going to figure out a profit equation and then maximize it.

00:11 So let's go for it.

00:12 Profit equation, that is p of x.

00:16 That's going to equal the r, which is 5x.

00:21 And then we're going to subtract this whole equation here, 0 .001x squared plus 1 .2x.

00:32 Plus 60.

00:34 Okay, so we're going to have to go ahead and make sure we distribute this across, so we can bring our light terms together minus 1 .2x minus 60.

00:50 I'm going to move this down a little bit.

00:56 Okay, so we bring our x terms together, and we're going to have 3 .8 minus this amount, minus 60.

01:12 Okay, so this is our p of x equation.

01:15 We can go ahead now and find the derivative of b of x.

01:22 And that's going to be 3 .8 minus 0 .002x minus 60.

01:35 Okay, so what happens here is we have the derivative.

01:41 And the derivative corresponds to if this were the equation, right, there's some sort of spot right at the top, that's the maximum.

01:49 It happens at sum of x.

01:51 We used p prime of x, right? that's the same as the sloping zero.

01:58 It's the slope.

01:59 So if p prime of x is zero, we know we have a flat spot, and so we know we have the maximum.

02:07 So we're going to set this equal to zero.

02:12 Ooh, you know what i just did? i just didn't have that minus 60th the end, but i did not mean to do that.

02:17 Bad, bad, bad...

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