a company sells one one of its product at rs.128 per unit in one year.Its cost function is given by X=0.01x^2+8x+7,where x represents the number of units.What will be the maximum profit of the company in nearest thousand rupee?
profit = income - cost
p(x) = 128x - (0.01x^2+8x+7) = -0.01x^2 + 120x - 7
Now just find the vertex in the usual way.
To find the maximum profit of the company, we need to determine the quantity of units that will maximize the profit. This can be done by finding the vertex of the cost function.
Given the cost function X = 0.01x^2 + 8x + 7, we can see that it is a quadratic function in the form of ax^2 + bx + c, where a = 0.01, b = 8, and c = 7.
To find the vertex of the quadratic equation, we will use the formula: x = -b / (2a).
Substituting the given values, we have: x = -8 / (2 * 0.01) = -8 / 0.02 = -8 / 2 = -4.
Note that the negative value of x obtained above indicates the number of units cannot be negative. Therefore, we disregard the negative value, and the number of units that will maximize the profit is x = 4.
To calculate the maximum profit, we need to subtract the cost from the revenue. The revenue is obtained by multiplying the selling price per unit (rs. 128) by the number of units sold (4).
Revenue = Selling Price * Number of Units = 128 * 4 = 512.
Now, to find the cost, we substitute x = 4 into the cost function X = 0.01x^2 + 8x + 7.
Cost = 0.01(4)^2 + 8(4) + 7
= 0.01(16) + 32 + 7
= 0.16 + 32 + 7
= 39.16.
Profit = Revenue - Cost = 512 - 39.16 = 472.84.
Rounding off the profit to the nearest thousand rupees gives us the maximum profit of the company, which is approximately 473 thousand rupees.
To find the maximum profit, we need to determine the number of units at which the profit is maximized.
The profit function can be calculated by subtracting the cost function from the revenue function.
Given that the selling price per unit is Rs. 128, the revenue function can be calculated as:
Revenue = Selling Price per unit × Number of Units
= 128x
The profit function can be calculated as:
Profit = Revenue - Cost
= 128x - (0.01x^2 + 8x + 7)
To find the maximum profit, we can take the derivative of the profit function with respect to x and set it equal to zero. Then, solve for x.
Let's calculate the derivative and solve for x:
Profit(x) = 128x - (0.01x^2 + 8x + 7)
Profit'(x) = 128 - 0.02x - 8
Setting Profit'(x) = 0, we have:
128 - 0.02x - 8 = 0
120 - 0.02x = 0
0.02x = 120
x = 120 / 0.02
x = 6000
So, the number of units at which the profit is maximized is 6000.
Now, we need to calculate the maximum profit by substituting the value of x into the profit function:
Profit = 128x - (0.01x^2 + 8x + 7)
Profit = 128 * 6000 - (0.01 * 6000^2 + 8 * 6000 + 7)
Profit = 768,000 - (360 + 48,000 + 7)
Profit ≈ 768,000 - 48,367
Profit ≈ 719,633
Therefore, the maximum profit of the company, rounded to the nearest thousand rupees, is Rs. 720,000.