Greg is opening a car wash. He estimates his cost equation as C= 5000 +0.05x and his revenue equation as where R=1.9x is the number of cars washed in a six-month period. Find the number of cars that must be washed in a six-month period for Greg to make a profit.
To break even, C=R
5000+0.05x = 1.9x
Isolate x and solve:
1.85x=5000
x=2702.7
So as of the 2703th car washed, he makes a profit.
To find the number of cars that must be washed in a six-month period for Greg to make a profit, we need to set the revenue equation equal to the cost equation and solve for x.
The revenue equation is R = 1.9x, where x is the number of cars washed in a six-month period.
The cost equation is C = 5000 + 0.05x.
To determine the breakeven point, set the revenue equal to the cost:
1.9x = 5000 + 0.05x
Subtract 0.05x from both sides:
1.85x = 5000
Divide both sides by 1.85:
x = 5000 / 1.85
Simplifying further,
x ≈ 2702.7
Therefore, Greg needs to wash approximately 2703 cars in a six-month period to make a profit.
To find the number of cars that must be washed in a six-month period for Greg to make a profit, we need to find when his revenue is greater than his costs.
Greg's cost equation is given as C = 5000 + 0.05x
And his revenue equation is given as R = 1.9x
To find when Greg makes a profit, we need to set the revenue equation greater than the cost equation and solve for x.
1.9x > 5000 + 0.05x
Let's solve for x:
1.9x - 0.05x > 5000
1.85x > 5000
To isolate x, divide both sides of the inequality by 1.85:
x > 5000 / 1.85
Calculating this, we get:
x > 2702.7
Since we can't have a fraction of a car washed, we round up to the nearest whole number:
x > 2703
Therefore, Greg must wash at least 2703 cars in a six-month period to make a profit.