Chapter 1 - Section 1.2 - Functions and Models - Exercises - Page 72: 21

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If $R(x)$ is the revenue from selling $x$ items at a price of m each, then $R$ is the linear function $R(x)=mx$ and the selling price $m$ can also be called the marginal revenue. Profit $=$ Revenue-Cost,$\qquad P(x)=R(x)-C(x)$ Breakeven occurs when $P=0$, or $R(x)=C(x)$. The break-even point is the number of items $x$ at which break even occurs. ----------------- Revenue function: $R(x)=100x,\quad (0 \leq x \leq 200)$ Profit function: $P(x)=100x - (2000+10x+0.2x^{2})$ $P(x)=-0.2x^{2}+90x-2000$ Breakeven: $P(x)=0$ $-0.2x^{2}+90x-2000=0\qquad /\times(-10)$ $2x^{2}-900x+20000=0\qquad /\div 2$ $x^{2}-450x+10000=0$ using the quadratic formula, $x=\displaystyle \frac{450\pm\sqrt{450^{2}-4(1)(10000)}}{2}$ $x=$23.4435562925 or $x=$426.556443707 We discard the higher value and conclude that the least (whole) number of jerseys that need to be sold so $P(x)$ is positive is 24