Im so lost on this...
Cannon Precision Instruments makes an automatic electronic flash. The estimated marginal profit associated with producing and selling theses electronic flashes is πβ²(π₯π₯) = β0.004π₯+ 20 dollars per unit per month when the production level is π₯π₯units per month. Cannonβs fixed cost for producing and selling these electronic flashes is $16,000/month. What is the maximum monthly profit when selling these electronic flashes? Round your answer to the nearest whole number and justify your solution by showing the appropriate Calculus work.
P(x) = -0.002x^2 + 20x + c
Since P'(0) = 20, and there are fixed costs of 16000, c must be 16020.
Better check my logic against what your text says.
To find the maximum monthly profit, we need to differentiate the profit function with respect to the production level, set the derivative equal to zero, and solve for the production level. Let's denote the monthly profit as π(π₯).
The profit for Cannon Precision Instruments is given by the formula:
π(π₯) = π
(π₯) - πΆ(π₯)
where π
(π₯) represents the revenue and πΆ(π₯) represents the cost.
The revenue is calculated as:
π
(π₯) = πβ²(π₯) Γ π₯
Substituting the given marginal profit function πβ²(π₯) = -0.004π₯ + 20, we have:
π
(π₯) = (-0.004π₯ + 20) Γ π₯
= -0.004π₯Β² + 20π₯
The cost is given as a fixed cost of $16,000/month, so:
πΆ(π₯) = $16,000
Now, we can calculate the profit function:
π(π₯) = (-0.004π₯Β² + 20π₯) - $16,000
To find the maximum monthly profit, we differentiate π(π₯) with respect to π₯:
πβ²(π₯) = (-0.008π₯ + 20)
Setting the derivative equal to zero:
-0.008π₯ + 20 = 0
Solving for π₯:
-0.008π₯ = -20
π₯ = (-20)/(-0.008)
π₯ = 2500
Therefore, the production level that maximizes monthly profit is 2500 units.
To calculate the actual maximum monthly profit, substitute π₯ = 2500 units into the profit function:
π(2500) = (-0.004(2500)Β² + 20(2500)) - $16,000
Simplify to find the maximum monthly profit.