Question

The total number of gallons of water in a tank at time t is modeled by the expression A(t)=30+8t-2/3(t+1)^3/2

Answer 1

193 gallons

Explanation:

Given

`A(t) = 30+8t-(2)/(3)(t+1)^{(3)/(2)}`

Required

Determine the maximum amount of water the tank can hold --- Missing from the question

Start by differentiating A w.r.t t

`A'(t) = 0 + 8 + (d)/(dt)[-(2)/(3)(t+1)^{(3)/(2)}]`

Solving:
`(d)/(dt)[-(2)/(3)(t+1)^{(3)/(2)}`

`(d)/(dt)[-(2)/(3)(t+1)^{(3)/(2)} = -(2)/(3)(d)/(dt)[(t+1)^{(3)/(2)}`

Apply power rule:

`(d)/(dt)[-(2)/(3)(t+1)^{(3)/(2)} = -(2)/(3)[(3)/(2)(t + 1)^{(3)/(2)-1} * (d)/(dt)[t+1]`

`(d)/(dt)[-(2)/(3)(t+1)^{(3)/(2)} = -(t + 1)^{(3)/(2)-1} * (d)/(dt)[t+1]`

`(d)/(dt)[-(2)/(3)(t+1)^{(3)/(2)} = -(t + 1)^{(3)/(2)-1} * [1+0]`

`(d)/(dt)[-(2)/(3)(t+1)^{(3)/(2)} = -(t + 1)^{(3)/(2)-1} * [1]`

`(d)/(dt)[-(2)/(3)(t+1)^{(3)/(2)} = -(t + 1)^{(3)/(2)-1}`

`(d)/(dt)[-(2)/(3)(t+1)^{(3)/(2)} = -(t + 1)^{(1)/(2)}`

So:

`A'(t) = 0 + 8 + (d)/(dt)[-(2)/(3)(t+1)^{(3)/(2)}]`

A'(t) = 0 + 8 + \frac{d}{dt}[-\frac{2}{3}(t+1)^{\frac{3}{2}}]

`A'(t) = 0 +8 -(t + 1)^{(1)/(2)}`

`A'(t) = 8 -(t + 1)^{(1)/(2)}`

Equate to 0 to solve for t

`A'(t) = 0`

`8 -(t + 1)^{(1)/(2)} = 0`

Collect Like Term

`-(t + 1)^{(1)/(2)} = -8`

Square both sides

`(-(t + 1)^{(1)/(2)})^2 = (-8)^2`

`t +1 = 64`

Make t the subject:

`t = 64 - 1`

`t = 63`

So, the tank is at maximum when t = 63.

Substitute 63 for t in:
`A(t) = 30+8t-(2)/(3)(t+1)^{(3)/(2)}`

`A(63) = 30+8*63-(2)/(3)(63+1)^{(3)/(2)}`

`A(63) = 30+8*63-(2)/(3)(64)^{(3)/(2)}`

`A(63) = 30+8*63-(2)/(3)*512}`

`A(63) = 30+8*63-(2*512)/(3)}`

`A(63) = 30+8*63-(1024)/(3)}`

`A(63) = 30+8*63-341.33`

`A(63) = 192.67`

Approximate:

`A(63) = 193`