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What is the maximum y-value of the following function?

$$y=8t - \frac{t^2}{2} -24 $$

It can be done by using the parabolic equation , setting the equation is equal zero. But is there any other straight forward or shortest method to do this?

4 Answers4

2

You can complete the square:

$$y=-\frac{1}{2}(t^2-16t+48)=-\frac{1}{2}((t-8)^2-16)=-\frac{1}{2}(t-8)^2+8$$

Since $-\dfrac{1}{2}(t-8)^2\le0$ for every $t$, the maximum occurs of $y$ when this expression is $0$ (which is at $t=8$). This gives the maximum of $y$ as $0+8=8$.

Peter Woolfitt
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2

for the form of $y = ax^2 + bx + c$ the min (or max) is :

$$y=c-\frac{b^2}{4a}$$

which in this case is :

$$-24+32=8$$

Shabbeh
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  • Is there a name for this theorem? Maybe a proof? I've never seen this before. – Cruncher Aug 07 '14 at 15:28
  • just find the min/max parametrically and substitute in the equation (again parametrically) – Shabbeh Aug 07 '14 at 15:34
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    I solved it after asking the question :). Did it using Aditya's answer, but keeping it general. – Cruncher Aug 07 '14 at 15:45
  • @Cruncher there's no theorem for this, per se. This fact can be derived by completing the square as described by Peter above, or by using calculus. – user2910265 Aug 07 '14 at 19:38
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Using calculus: $$y'=8-t$$ Using derivative test $8-t=0\implies t=8$ So,$$y_{\text{extremum}}=64-32-24=8$$ Second derivate: $$y''=-1<0$$ So extremum is a maximum.

RE60K
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0

$$y(t)=8t-\frac{t^2}{2}-24$$

$$y'(t)=0 \Rightarrow 8-t=0 \Rightarrow t=8$$

For $t>8:$ $y>0$

For $t<8:$ $y<0$

So, $y$ is decreasing on $(-\infty, 0]$ and increasing on $[0,+\infty)$.

Therefore, $y$ achieves its minimum at $t=8$, which is equal to $y(8)=8$.

Mary Star
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