Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.8 - Inequalities - 1.8 Exercises - Page 91: 122

Answer

$400\leq x \leq 4400$

Work Step by Step

We have to find interval of the following inequality: $Profit \geq 2400$ $Profit = revenue - cost$ $=$ $20x-(2000+8x+0.0025x^2)$ Let the $Profit$ be $P$. $P=20x-2000-8x-0.0025x^2=-0.0025x^2+12x-2000$ So, we have to find interval of the inequality: $-0.0025x^2+12x-2000\geq 2400$ $-0.0025x^2+12x-4400\geq 0$ (Multiply by -400) $x^2-4800x+1760000\leq0$ Now, we can write it as a quadratic equation and solve it using interval notation $x^2-4800x+1760000=0$ $D=b^2-4ac=16,000,000$ $x=\frac{-b±\sqrt{D}}{2a}=\frac{4800±4000}{2}$ $x_1=400$ ; $x_2=4400$ We have $3$ possible intervals: $(-\infty, 400]$ - Positive $[400, 4400]$ - Negative $[4400, +\infty)$ - Positive Now we get back to the last inequality. We needed interval less than or equal to $0$, so the second interval is appropriate: $[400, 4400]$ $400\leq x \leq 4400$
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