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# Find two numbers whose sum is 24 and whose product is as large as possible.

Toolbox:
• $\large\frac{d}{dx}\big(x^n)=nx^{n-1}$
• For maximum value $\large\frac{dP}{dx}$$=0 Step 1: Given Sum of two no\rightarrow 24 Product \rightarrow as large as possible Let the required number be x and (24-x) Their product P=x(24-x) \qquad\qquad\quad\;\;\;=24x-x^2 Step 2: For maximum value Now \large\frac{dP}{dx}$$=0$
$P=24x-x^2$
Differentiating with respect to x we get
$\large\frac{dP}{dx}$$=24-2x 24-2x=0 -2x=-24 x=12 \large\frac{d^2P}{dx^2}$$=-2 < 0$
$\Rightarrow P$ is minimum at $x=12$
Hence the required numbers are 12 and $(24-12)\Rightarrow (i.e)12$
edited Aug 19, 2013