Question

A juice company has found that the marginal cost of producing x pints of fresh-squeezed orange juice is given by the function below, where c'(x) is in dollars. Approximate the total cost of producing 261 pt of juice. using 3 subintervals over [0,261] and the left endpoint of each subinterval. C'(x) = 0.000006x-0.003x+5, for x S 350 The total cost is about $ (Round the final answer to the nearest cent as needed. Round all intermediate values to the nearest thousandth as needed.)

Answer 1

To approximate the total cost of producing 261 pt of juice, use the left endpoint approximation with 3 subintervals. Plug in the values of the marginal cost function for each subinterval and calculate the sum to find the approximate total cost.

Step-by-step explanation:

To approximate the total cost of producing 261 pints of juice using 3 subintervals over [0,261] and the left endpoint of each subinterval, we can approximate the integral of the marginal cost function over the interval [0,261].

First, we need to find the values of c'(x) for x = 0, 87, 174, and 261 using the given marginal cost function:

c'(x) = 0.000006x^2 - 0.003x + 5

Plugging in the values, we get:

c'(0) = 5

c'(87) = 0.000006(87)^2 - 0.003(87) + 5

c'(174) = 0.000006(174)^2 - 0.003(174) + 5

c'(261) = 0.000006(261)^2 - 0.003(261) + 5

Next, we can use the left endpoint approximation formula to approximate the integral:

Total cost ≈ c'(0)(87-0) + c'(87)(174-87) + c'(174)(261-174)

Calculating this expression will give us the approximate total cost of producing 261 pints of juice.

Answer 2

The approximate total cost of producing 261 pints of juice using three subintervals and the left endpoint of each subinterval is approximately $1562.51.

To find the total cost of producing 261 pints of juice using three subintervals and the left endpoint of each subinterval, you'll need to use the definite integral of the marginal cost function over the given range.

The marginal cost function
`\(c'(x)\)` is given as:

`\[c'(x) = 0.000006x^2 - 0.003x + 5\]`

Let's find the definite integral of this function over the interval [0, 261] using three subintervals:

`\[c(x) = \int c'(x) \,dx = \int (0.000006x^2 - 0.003x + 5) \,dx\]`

First, integrate each term separately:

`\[c(x) = 0.000002x^3 - 0.0015x^2 + 5x + C\]`

Now, evaluate
`\(c(x)\)` at the upper and lower limits of integration:

Using the upper limit (261):

`\[c(261) = 0.000002 * 261^3 - 0.0015 * 261^2 + 5 * 261 + C\]`

`\[c(261) \approx 0.000002 * 179,847,261 - 0.0015 * 68,121 + 1,305 + C\]`

`\[c(261) \approx 359.694 - 102.182 + 1,305 + C\]`

`\[c(261) \approx 1,562.512 + C\]`

Using the lower limit (0):

`\[c(0) = 0.000002 * 0^3 - 0.0015 * 0^2 + 5 * 0 + C\]`

`\[c(0) = 0 - 0 + 0 + C\]`

`\[c(0) = C\]`

Now, subtract the lower limit value from the upper limit value to find the total cost:

Total cost
`\(= c(261) - c(0) = 1,562.512 + C - C = 1,562.512\)`

Therefore, The answer is $1562.51.