# The Mechanical Universe- Chapter 3- question 6

Suppose your income $$y$$ is directly proportional to the number $$x$$ of hours you work: $$y = cx$$, where $$c$$ is a constant. In addition, suppose you're a big spender and the money $$z$$ you spend varies with income as $$z(y) = a + by^2$$, where $$a$$ and $$b$$ are also constants. For your income to be larger than your expenses, what conditions on $$y$$, $$a$$, and $$b$$ must be met?

I'm not entirely sure how I am supposed to write the answers(There is no solution on the back of the book). I know that $$a < 0$$ and $$|a/y^2|>b$$ but is that how I am supposed to express the conditions? Or am I supposed to express the conditions in terms of $$c$$ and $$x$$?

$$y\ge z\\ y\ge a+by^2\\ 0\ge by^2 - y + a$$
if $$a,b > 0$$
$$\frac {1 - \sqrt{1-4ab}}{2} \le y \le\frac {1 + \sqrt{1-4ab}}{2}$$
and $$1-4ab \ge 0$$
The income is $$y$$, the expenses is $$a+by^2$$, so the condition income greater than expenses is $$y>a+by^2$$ You can rewrite this as $$by^2-y+a<0$$ Calculate the roots of $$by^2-y+a=0$$. If $$b>0$$, then $$y$$ has to be between the roots. if $$b<0$$, $y has to be on the outside. • In your second inequality, it should be$\lt 0$, not$\gt 0\$. – John Omielan Jul 16 '19 at 22:22