To solve the problem of finding the maximum or minimum of the function [tex]\( y = -16x^2 + 64x + 80 \)[/tex], we can follow these detailed steps:
1. Rewrite the problem for clarity:
We begin by writing the given function explicitly as [tex]\( y = -16x^2 + 64x + 80 \)[/tex].
2. Find the critical points:
To find the critical points, we need to determine where the derivative of the function equals zero. First, compute the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[
\frac{dy}{dx} = \frac{d}{dx}(-16x^2 + 64x + 80)
\][/tex]
Upon differentiating, we obtain:
[tex]\[
\frac{dy}{dx} = -32x + 64
\][/tex]
Set this derivative equal to zero to find the critical points:
[tex]\[
-32x + 64 = 0
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
-32x = -64 \implies x = 2
\][/tex]
3. Determine the nature of the critical points:
To determine whether the critical point is a maximum or a minimum, we evaluate the second derivative of the function. First, compute the second derivative:
[tex]\[
\frac{d^2y}{dx^2} = \frac{d}{dx}(-32x + 64)
\][/tex]
Upon differentiating, we obtain:
[tex]\[
\frac{d^2y}{dx^2} = -32
\][/tex]
Since the second derivative is negative ([tex]\(-32 < 0\)[/tex]), this indicates that the function [tex]\( y \)[/tex] has a maximum at [tex]\( x = 2 \)[/tex].
4. Evaluate the function at the critical point:
To find the maximum value of the function, substitute [tex]\( x = 2 \)[/tex] back into the original function:
[tex]\[
y = -16(2)^2 + 64(2) + 80
\][/tex]
Calculating this, we get:
[tex]\[
y = -16(4) + 128 + 80
\][/tex]
[tex]\[
y = -64 + 128 + 80
\][/tex]
[tex]\[
y = 144
\][/tex]
Therefore, the function [tex]\( y = -16x^2 + 64x + 80 \)[/tex] has a maximum value at [tex]\( x = 2 \)[/tex] and that maximum value is [tex]\( y = 144 \)[/tex].
Thus, the maximum point is [tex]\((2, 144)\)[/tex].
