Answer:
[tex]b = 10[/tex]
The smallest number of boats that must be produced to make a profit of $ 75,000 is 10 boats
Step-by-step explanation:
We have this equation
[tex]P(b) = -25b^3 + 750b^2 + 2500b[/tex]
This equation shows the income obtained based on the number of boats sold. We want to know the minimum number of boats that must be sold to obtain $ 75 000.
Then we must equalize the equation to 75 000 and clear b.
3 solutions will be obtained (because it is a polynomial of degree 3), and the lowest value will be taken.
[tex]P(b) = 75000[/tex]
[tex]-25b^3 + 750b^2 + 2500b = 75000[/tex]
Now we need to solve the equation, for that we seek to factor the polynomial
[tex]-25b^3 + 750b^2 + 2500b - 75000 = 0[/tex]
[tex]b^3 -30b^2 - 100b + 3000 = 0[/tex] We divide the equation by -25
[tex]b^2(b-30) - 100(b-30) = 0[/tex] We take common factor b
[tex](b^2 -100)(b-30) = 0[/tex] We take out common factor ([tex]b-30[/tex])
Finally the solutions are:
[tex]b = 30\\\\b^2 -100 = 0\\\\b^2 = 100[/tex]
[tex]b = 10[/tex] and [tex]b = -10[/tex]
We take the least positive solution (because you can not produce -10 boats)
Finally we have that [tex]b = 10[/tex]
