1. Profit = Sale price - Cost
Part (a)
Profit = Sale price - Cost
Profit = 120x – ∫ 6√x 25
Profit = 120(25) – 4[x3/2] 25
Profit = $2500

2. Part (b)
This represents the cost, in dollars, of building the last five meters of a 30 meter cable.

3. Profit = Sale price - Cost
Part (c)
Profit = Sale price - Cost
Profit = 120k – ∫6√x k OR
Profit = 120k – 4k3/2
P(k) = 4k(30 – √k)

4. Part (d)
Profit is positive on the interval 0 < k < 900, meaning a maximum occurs on this interval when the derivative is zero.
From (c)…
P(k) = 120k – 4k3/2
P’(k) = 120 – 6√k
0 = 120 – 6√k

5. Part (d)
0 = 120 – 6√k
6√k = 120
√k = 20
k = 400
Mighty earns maximum profit when they sell a 400 meter cable. Now we have to calculate the profit of the cable.

6. Part (d)
P(400) = 120(400) – 4(400)3/2
P(400) = $16,000