You can put this solution on YOUR website! Find the maximum value of the equation: y=-16t^2+80t+2
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y=-16t^2+80t+2
y' = -32t + 80 = 0
t = 2.5
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y@(t=2.5) = -16*2.5^2 80*2.5 + 2
= -100 + 200 + 2
= 102
You can put this solution on YOUR website! Given
(1) -16t^2+80t+2 = 0 which is of the standard quadratic form
(2) ax^2 + bx + c = 0
If we set (2) equal to f(x) and let
(3) y = f(x) we get
(4) y = ax^2 + bx + c which is also called the formula for a parabola.
You may have learned that the x-axis of symmetry for the parabola is given by
(5) x = -b/2a
In (1) we have a = -16 and b = 80, so we get
(6) x-axis of symmetry = -(80)/(2*(-16)) or
(7) x-axis of symmetry = 80/32 or
(8) x-axis of symmetry = 5/2
In your problem x = t so we get
(9) t = 5/2
Another property of the parabola is the it has a maximum or minimum when x (or t in this case) is at the x value of the x-axis of symmetry. Therefore set t=5/2 in (1) to get the maximum or
(10) Max = -16(5/2)^2+80(5/2)+2 or
(11) Max = -16*25/4 + 80*5/2 + 2 or
(12) Max = -100 +200 +2 or
(13) Max = 102
Answer: The given equation has a maximum value of 102 when t is 5/2.
PS If you are into calculus, you can get (5) by taking the derivative of (4) with respect to x, set this derivative equal to zero and solve for x, or
(14) 2*a*x + b = 0 or
(15) x = -b/2a
All the derivative is, is the slope of the quadratic, and (15) tells us when the slope is zero, i.e a peak or valley.
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