Sound Software estimates that it will sell N units of a...

Sound Software estimates that it will sell N units of a program after spending a dollars on advertising, where $N(a)=-a^{2}+300 a+6, \quad 0 \leq a \leq 300$ and $a$ is in thousands of dollars. Find the maximum number of units that can be sold and the amount that must be spent on advertising in order to achieve that maximum.

Sound Software estimates that it will sell N units of a program after spending a dollars on advertising, where $N(a)=-a^{2}+300 a+6, \quad 0 \leq a \leq 300$
and $a$ is in thousands of dollars. Find the maximum number of units that can be sold and the amount that must be spent on advertising in order to achieve that maximum.

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Key Concepts

Optimization

Optimization involves finding the best or most efficient solution in a given problem, such as maximizing profits or minimizing costs. In the context of quadratic functions, optimization is achieved by calculating the vertex, which indicates the maximum or minimum value of the function within a specified domain.

Vertex of a Parabola

The vertex of a parabola is the point where the function reaches its maximum or minimum value, depending on whether the quadratic opens downward or upward. For a quadratic function in the form ax² + bx + c, the vertex is found using the formula x = -b/(2a), and this point is critical in determining the optimal outcome in problems involving quadratic relationships.

Quadratic Functions

Quadratic functions are polynomial expressions of degree two that can be represented in the standard form f(x) = ax² + bx + c. They are often used to model real-world situations where the relationship between variables is parabolic, displaying a distinct maximum or minimum value under certain conditions.

Transcript

00:01 Given this function here, n of a is equal to negative a square plus 300a plus 6, where i say, well, if not t, where a, right, is between 0 and 300.

00:20 So, right, where a is between, so where a is between 0 and 300.

00:25 So i wrote t, right, but where a is between, so where a is between 0 ,000.

00:34 In 300.

00:35 Okay.

00:37 So where n here is the units of the program, and a is the amount in thousands of dollars.

00:44 So first we find our derivative of our function.

00:47 So that would be n prime of a.

00:51 So just take the derivative here piece by piece and see that our derivative is equal to negative to a plus 300.

01:05 So there's our derivative.

01:06 We're going to find out clinical values.

01:08 We take our derivative and we set it equal to zero.

01:11 So we have negative 2a.

01:15 Well, plus 300 is equal to 0 implies that negative 2a is equal to negative 300.

01:24 Okay, and then we divide both sides through here by negative 2 to get that a is equal to 150.

01:31 So our critical value is a being equal to 150.

01:35 So then we list out our critical values and endpoints.

01:38 So we have our endpoint of zero, our critical value of 150, and our other endpoints of 300.

01:45 So then we go back and we evaluate our original function at these values...

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