Respuesta :

carlosego
For this case we have the following function:
 [tex]f (x) = 3ax-x ^ 2 [/tex]
 To find the maximum of the function, what we should do is to derive the equation.
 We have then:
 [tex]f '(x) = 3a-2x [/tex]
 We match zero:
 [tex]3a-2x = 0 [/tex]
 We clear the value of x:
 [tex]2x = 3a x = (2/3) a[/tex]
 We substitute the value of x in the function to find the maximum:
 [tex]f ((2/3) a) = 3a ((2/3) a) - ((2/3) a) ^ 2 [/tex]
 Rewriting:
 [tex]f ((2/3) a) = 2a ^ 2 - (4/9) a ^ 2 f ((2/3) a) = (18/9) a ^ 2 - (4/9) a ^ 2 f ((2/3) a) = (14/9) a ^ 2[/tex]
 Answer:
 a formula in terms of a for the maximum of f (x) is:
 [tex]f ((2/3) a) = (14/9) a ^ 2[/tex]

Using the vertex of the quadratic equation, it is found that the formula in terms of a for the maximum of f(x) is of:

[tex]f_{MAX} = \frac{9a^2}{4}[/tex]

What is the vertex of a quadratic equation?

A quadratic equation is modeled by:

[tex]y = ax^2 + bx + c[/tex]

The vertex is given by:

[tex](x_v, y_v)[/tex]

In which:

[tex]x_v = -\frac{b}{2a}[/tex]

[tex]y_v = -\frac{b^2 - 4ac}{4a}[/tex]

Considering the coefficient a, we have that:

  • If a < 0, the vertex is a maximum point.
  • If a > 0, the vertex is a minimum point.

In this problem, the function is:

[tex]f(x) = 3ax - x^2[/tex]

The coefficients are a = -1, b = 3a, c = 0, hence:

[tex]f_{MAX} = -\frac{(3a)^2}{-4}[/tex]

[tex]f_{MAX} = \frac{9a^2}{4}[/tex]

More can be learned about the vertex of a quadratic equation at https://brainly.com/question/24737967

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