I've been trying to solve this problem for the past 45 minutes and could really use some help. I'm given that water flows into a tank at a rate of (5t+4) gallons/min. Water flows out of the tank at a rate of 0.5(t^2) gallons/min. At t=0 min the tank contains 100 gallons.

The questions are: 1) Write an expression for the amount of water in the tank at any given time t
2) When will the quantity of water in the tank be a maximum and how much is the quantity of water at the time? Show the work that leads to your answers.
3) Find the average number of gallons of water in the tank in the first five minutes, and determine the time during the first 5 minutes when this average number is actually obtained. Justify your answers.

What I have so far (and I'm pretty sure it's worng) is 1) V(t)= -0.5t^2+5t+104
2) V'(t)=-t + 5 solve for V'(0) and got 5.
3) Solved for V(5) and plugged that answer back into V'(t). So, I got V(5)= 116.5 and then plugged it in to get V'(116.5)= -111.5

V(t) = 100 + (5t+4) - t^2/2

so you are correct there.

V'(t) = 5-t
So, you are correct that V is a max at t=5

The average value of a function is
(∫[a,b] f(t) dt)/(b-a) So, that means that #3 is

∫[0,5] (-0.5t^2+5t+104) dt = 337/3

So, you want to find when

(-t^2/2 + 5t + 104) = 337/3
t = 5 - 5/√3 ≈ 2.113

To help you solve this problem, let's break it down step by step:

1) Writing an expression for the amount of water in the tank at any given time t:

To find the amount of water in the tank at any given time t, you need to add up the amount of water that flows into the tank and subtract the amount of water that flows out of the tank.

Given that water flows into the tank at a rate of (5t+4) gallons/min and water flows out of the tank at a rate of 0.5(t^2) gallons/min, the expression for the amount of water in the tank at time t can be written as:

V(t) = integral[(5t+4) - 0.5(t^2)] dt

To evaluate this integral, you can expand the expression and integrate each term separately. The integral of (5t+4) is (5/2)t^2 + 4t, and the integral of 0.5(t^2) is (1/6)t^3. Combining these terms, the expression for the amount of water in the tank is:

V(t) = (5/2)t^2 + 4t - (1/6)t^3 + C

Where C is a constant that represents the initial amount of water in the tank, which is given as 100 gallons when t=0. So, we can rewrite the expression as:

V(t) = (5/2)t^2 + 4t - (1/6)t^3 + 100

2) Finding when the quantity of water in the tank is a maximum and the corresponding quantity of water at that time:

To find the maximum quantity of water in the tank, we need to find the time at which the derivative of the volume function V(t) is equal to zero.

V'(t) = (5t) - (1/2)t^2

Setting V'(t) to zero and solving for t:

0 = 5t - (1/2)t^2

Multiplying by 2 to remove the fraction:

0 = 10t - t^2

Rearranging the equation:

t^2 - 10t = 0

Factoring:

t(t - 10) = 0

Solving for t, we have two solutions: t=0 and t=10. However, t=0 is not a valid time since it corresponds to the initial condition. So the only valid solution is t=10.

To find the corresponding quantity of water at t=10, we can substitute t=10 into the volume function:

V(10) = (5/2)(10^2) + 4(10) - (1/6)(10^3) + 100

V(10) = 250 + 40 - (1000/6) + 100

V(10) = 150 + 40 - 166.67 + 100

V(10) ≈ 123.33 gallons

Therefore, the maximum quantity of water in the tank occurs at t=10 minutes, and the quantity of water at that time is approximately 123.33 gallons.

3) Finding the average number of gallons of water in the tank in the first five minutes and determining the time during the first 5 minutes when this average number is actually obtained:

The average number of gallons of water in the tank in the first five minutes can be found by calculating the average value of the volume function V(t) over the interval t = [0, 5].

Average = (1/(5-0)) * integral[V(t)] from t = 0 to 5

To find the time during the first 5 minutes when this average number is actually obtained, we need to find the time within this interval where the volume function V(t) equals the average value. This can be done by solving the equation V(t) = Average.

To calculate these values, first, we need to evaluate the integral of V(t) from t = 0 to 5:

Integral[V(t)] from t = 0 to 5 = integral[(5/2)t^2 + 4t - (1/6)t^3 + 100] dt from t=0 to 5

Evaluating this integral, we can substitute the limits of integration:

Integral[V(t)] from t = 0 to 5 = [(5/6)(5^3) + 2(5^2) - (1/24)(5^4) + 100(5)] - 100(0)

Integral[V(t)] from t = 0 to 5 ≈ 625 + 50 - 104.17 + 500

Integral[V(t)] from t = 0 to 5 ≈ 1070.83 gallons

Now, let's find the average:

Average = (1/(5-0)) * Integral[V(t)] from t = 0 to 5

Average = (1/5) * 1070.83

Average ≈ 214.17 gallons

To find the time during the first 5 minutes when this average number is actually obtained, we need to solve the equation V(t) = Average.

V(t) = (5/2)t^2 + 4t - (1/6)t^3 + 100

(5/2)t^2 + 4t - (1/6)t^3 + 100 = 214.17

Simplifying the equation and solving it is a bit challenging algebraically, so you can use numerical methods or graphing calculators to find the approximate solution.