Answer:
a) 66.25
b) 100
c) 83.75
Explanation:
We are given
![W(t)=-3.75t^2+30t+40](https://img.qammunity.org/2021/formulas/mathematics/college/zpy4yra5r44a44rv50fe991ozw72hpr1eo.png)
a)
![W(0) = -3.75+30+40=66.25](https://img.qammunity.org/2021/formulas/mathematics/college/7wxs3xh203y03qmz9qgxif35vpng3mnkrc.png)
b) Maximum value can be found by taking derivative of W(t) with respect to t.
![(dW(t))/(dt) =-7.5t+30=0](https://img.qammunity.org/2021/formulas/mathematics/college/zzf284oym8ggo831r2nf0b5pxq5d02y8zu.png)
So, t = 4 is absolute maximum.
Thus, maximum value of W(t) is occured at t = 4.
![-3.75*4^2+30*4+40=100](https://img.qammunity.org/2021/formulas/mathematics/college/ifof6cvcokpgn3bsdl3bw78gah42dwwsih.png)
c) Average value can be found as follows,
![W_(avg)=(1)/(5-0) \int\limits^5_0 (-3.75t^2+30t+40)dt=\\\\=(1)/(5) (-1.25t^3+15t^2+40t)|^5_0=(418.75)/(5) =83.75](https://img.qammunity.org/2021/formulas/mathematics/college/lzcs9k8hmj1gjcn9ynzj52k3ksxiycqyoi.png)