Question

model the annual revenue as R(t)=350(39t+60)e −0.2tmillion dollars. (a) Determine when to the nearest 0.1 year the revenue was projected to peak. (Submit your answer in terms of t.) o yr (b) Determine the revenue, to the nearest $1 million, at that time. ⩽ & million

Answer 1

At (t ≈ 3.47) years, the revenue peaks. At this time, the revenue is approximately \$47.3 million, derived from (R(3.47) ≈ 47332.5) million dollars.

Given:

`\(R(t) = 350(39t + 60)e^(-0.2t)\)`million dollars

(a) To determine when the revenue peaks, find
`\(t\)` when
`\(R'(t) = 0\)`, where
`\(R'(t)\)`is the derivative of
`\(R(t)\)`.

The derivative of
`\(R(t)\)` with respect to
`\(t\)` can be found using the product rule and chain rule:

Given
`\(R(t) = 350(39t + 60)e^(-0.2t)\)`,

`\[ R'(t) = 350 \cdot (39)e^(-0.2t) + 350(39t + 60) \cdot (-0.2)e^(-0.2t) \]`

Setting
`\(R'(t)\)` to zero to find the critical points:

`\[ 350 \cdot (39)e^(-0.2t) + 350(39t + 60) \cdot (-0.2)e^(-0.2t) = 0 \]`

Factor out
`\(e^(-0.2t)\)`:

`\[ e^(-0.2t)(350 \cdot 39 - 350(39t + 60) \cdot 0.2) = 0 \]`

Solve for
`\(t\)`:

`\[ 350 \cdot 39 - 350(39t + 60) \cdot 0.2 = 0 \]`

`\[ 13650 - 70(39t + 60) = 0 \]`

`\[ 13650 - 2730t - 4200 = 0 \]`

`\[ 9450 - 2730t = 0 \]`

`\[ 2730t = 9450 \]`

`\[ t = (9450)/(2730) \]`

t ≈ 3.47 { years (to the nearest 0.1 year)}

(b) To find the revenue at \(t ≈ 3.47\) years, substitute this value into \(R(t)\):

`\[ R(3.47) = 350(39 \cdot 3.47 + 60)e^(-0.2 \cdot 3.47) \]`

`\[ R(3.47) = 350(135.33)e^(-0.694) \]`

R(3.47) ≈ 47332.5 { million dollars (to the nearest million)}

Therefore, at \(t ≈ 3.47\) years, the revenue is approximately \$47.3 million.

the completed question:

Model the annual revenue as \(R(t) = 350(39t+60)e^{-0.2t}\) million dollars.

(a) Calculate, to the nearest 0.1 year, when the revenue was projected to peak. Provide your answer in terms of \(t\).

(b) Calculate the revenue, rounded to the nearest \$1 million, at that time. Submit your answer in millions of dollars.