A factory produces x calculators per day. The total daily cost in Shillings incured is 5x^2-800x+500. If the calculators are sold for sh (100-10x) each,
i, Determine the profit function
ii, Find the number of calculators that would maximize the daily profit
iii, What is the daily maximum profit?
A factory produces x calculators per day. The total daily cost in Shillings incured is 5x^2-800x+500. If the calculators are sold for sh (100-10x) each,
i, Determine the profit function
ii, Find the number of calculators that would maximize the daily profit
iii, What is the daily maximum profit?
To find the profit function, we need to subtract the cost from the revenue.
i. Profit function:
Profit = Revenue - Cost
Revenue = Number of calculators sold * Selling price per calculator
Revenue = x * (100 - 10x) = 100x - 10x^2
Cost = 5x^2 - 800x + 500
Profit = Revenue - Cost
Profit = 100x - 10x^2 - (5x^2 - 800x + 500)
Profit = 100x - 10x^2 - 5x^2 + 800x - 500
Profit = -15x^2 + 900x - 500
Therefore, the profit function is given by: Profit = -15x^2 + 900x - 500.
ii. To find the number of calculators that would maximize the daily profit, we need to find the vertex of the quadratic function. The x-coordinate of the vertex can be found using the formula: x = -b / (2a).
In this case, a = -15 and b = 900.
x = -900 / (2*(-15))
x = -900 / (-30)
x = 30
The number of calculators that would maximize the daily profit is 30.
iii. To find the maximum daily profit, substitute the value of x found in step ii into the profit function:
Profit = -15x^2 + 900x - 500
Profit = -15*(30)^2 + 900*(30) - 500
Profit = -15*900 + 27000 - 500
Profit = -13500 + 27000 - 500
Profit = 13500 - 500
Profit = 13000 shillings
Therefore, the maximum daily profit is 13,000 shillings.