The revenue and cost equations for a product are R=x (50-0.002x) and C= 12x+ 150000. Where R and C are measured in dollars and x represents the number of units sold. How many units must be sold to get a profit of at least 1,650,000
well, since profit is revenue less cost,
P(x) = R(x)-C(x)
you just need to solve for x in
0.002x^2 + 38x - 150000 >= 1650000
To find the number of units that must be sold to achieve a profit of at least $1,650,000, we need to determine the point of intersection between the revenue (R) and cost (C) equations.
Profit (P) is calculated by subtracting the cost from the revenue:
P = R - C
We know that we want a profit of at least $1,650,000, so we can set up the following equation:
P ≥ 1,650,000
Substituting the revenue and cost equations into the profit equation, we get:
x (50 - 0.002x) - (12x + 150,000) ≥ 1,650,000
Expanding and rearranging the equation, we have:
50x - 0.002x^2 - 12x - 150,000 ≥ 1,650,000
Combining like terms, we have the following quadratic equation:
-0.002x^2 + 38x - 1,800,000 ≥ 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -0.002, b = 38, and c = -1,800,000.
Plugging these values into the quadratic formula, we get:
x = (-38 ± √(38^2 - 4(-0.002)(-1,800,000))) / (2(-0.002))
Simplifying further, we have:
x = (-38 ± √(1444 - 14400)) / (-0.004)
x = (-38 ± √(12956)) / (-0.004)
x = (-38 ± 113.86) / (-0.004)
Now, we need to consider both possibilities for x:
1) x = (-38 + 113.86) / (-0.004)
x = 75.86 / (-0.004)
x ≈ -18965
2) x = (-38 - 113.86) / (-0.004)
x = -151.86 / (-0.004)
x ≈ 37965
Since the number of units sold cannot be negative (as it represents physical quantities), we ignore the first solution.
Therefore, the minimum number of units that must be sold to achieve a profit of at least $1,650,000 is approximately 37,965 units.