## Calculus 8th Edition

$(a,b) = (50, -50)$
Find two numbers whose difference is 100 and whose product is a minimum. Write out the equations $a-b=100$ and $y=ab$ Substitute $a=100+b$ $y=(100+b)b$ $y=100b + b^2$ The problem asks to optimize y, so differentiate and set the derivative $=0$ $y'=100+2b$ $-100=2b$ $b=-50$ Use the second derivative test to confirm that $b=-50$ is in fact a minimum. $y''= 2$ Concave up on a critical point confirms that $b=-50$ is a minimum. Therefore, $a=50$. $(a,b) = (50, -50)$