Given:
The amount of profit, y, made by the company, is related to the selling price of each widget, x, by the given equation
[tex]y=-29x^2+1388x-10040[/tex]
To find:
The maximum amount of profit the company can make, to the nearest dollar.
Solution:
If a quadratic equation is [tex]f(x)=ax^2+bx+c[/tex], then the vertex is
[tex]Vertex=\left(-\dfrac{b}{2a},f(-\dfrac{b}{2a})\right)[/tex]
If a>0, then vertex is the minimum point and if a<0, then the vertex is the maximum point.
We have,
[tex]y=-29x^2+1388x-10040[/tex]
Here, [tex]a=-29,b=1388, c=-10040[/tex]. Clearly, a<0. So, the vertex is the point of maxima.
[tex]-\dfrac{b}{2a}=-\dfrac{1388}{2(-29)}[/tex]
[tex]-\dfrac{b}{2a}=-\dfrac{1388}{-58}[/tex]
[tex]-\dfrac{b}{2a}\approx 23.931[/tex]
Putting x=23.931 in the given equation, we get
[tex]y=-29(23.931)^2+1388(23.931)-10040[/tex]
[tex]y=-16608.09+33216.228-10040[/tex]
[tex]y=6568.138[/tex]
The vertex is at (23.931,6568.138).
Therefore, the maximum profit is $6568.138 when x=23.931.
