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?
P(x) = R(x)-C(x) = x(100-10x) - (5x^2-800x+500)
That' just a parabola, so its vertex is easy to find, no?
To determine the profit function, we need to subtract the cost from the revenue.
i. Profit function (P) = Revenue - Cost
The revenue obtained from selling x calculators at a price of sh (100-10x) each is given by R = x(100-10x) = 100x - 10x^2.
The daily cost incurred is 5x^2 - 800x + 500.
So, the profit function is: P = R - Cost = (100x - 10x^2) - (5x^2 - 800x + 500).
Simplifying, we have: P = 100x - 10x^2 - 5x^2 + 800x - 500.
Combining like terms: P = -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 profit function. The x-coordinate of the vertex is the value that maximizes the profit.
The x-coordinate of the vertex can be found using the formula: x = -b/2a, where a, b, and c are the coefficients of the quadratic equation.
For the profit function P = -15x^2 + 900x - 500, the coefficient of x^2 is -15, and the coefficient of x is 900.
Using the formula, we have: x = -900/(2*(-15)) = -900/(-30) = 30.
Therefore, the number of calculators that would maximize the daily profit is 30.
iii. To find the maximum daily profit, substitute the value of x (30) into the profit function:
P = -15(30)^2 + 900(30) - 500.
Simplifying, we have: P = -15(900) + 27000 - 500 = -13500 + 27000 - 500 = 13500.
Therefore, the daily maximum profit is 13,500 Shillings.