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Part (a)

Part (a). 25. Profit = 120x – ∫ 6 √ x. 0. 25. Profit = 120(25) – 4[x 3/2 ]. 0. Profit = Sale price - Cost. Profit = $2500. Part (b). This represents the cost, in dollars, of building the last five meters of a 30 meter cable. Part (c). k. Profit = 120k – ∫ 6 √ x. 0.

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Part (a)

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  1. Part (a) 25 Profit = 120x – ∫ 6√x 0 25 Profit = 120(25) – 4[x3/2] 0 Profit = Sale price - Cost Profit = $2500

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

  3. Part (c) k Profit = 120k – ∫6√x 0 P(k) = 4k(30 – √k) Profit = Sale price - Cost OR Profit = 120k – 4k3/2

  4. Part (d) P’(k) = 120 – 6√k 0 = 120 – 6√k 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

  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

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