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If $\mathrm{x}+\mathrm{y}=12$, what is the maximum value of $\mathrm{xy}$ ?
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The correct answer is:
36
Given $x+y=12$ $\mathrm{y}=12-\mathrm{x}$
so, $x y=x(12-x)=12 x-x^{2}$
Let $f(x)=12 x-x^{2}$
$\mathrm{f}^{\prime}(\mathrm{x})=12-2 \mathrm{x}$
To get maximum or minimum value $\mathrm{f}^{\prime}(\mathrm{x})=0$ and $\mathrm{f}^{\prime \prime}(\mathrm{x}) < 0$ it is maximum
$\mathrm{f}^{\prime \prime}(\mathrm{x})=-2 < 0$
so, $\mathrm{f}^{\prime}(\mathrm{x})=0$ will give maximum value. so, $12-2 x=0 \Rightarrow x=6$ and $x+y=12 \Rightarrow y=6$
Hence, $\mathrm{y}=6$ and $\mathrm{f}(\mathrm{x})=12 \mathrm{x}-\mathrm{x}^{2}=12 \times 6-36=36$
so, $x y=x(12-x)=12 x-x^{2}$
Let $f(x)=12 x-x^{2}$
$\mathrm{f}^{\prime}(\mathrm{x})=12-2 \mathrm{x}$
To get maximum or minimum value $\mathrm{f}^{\prime}(\mathrm{x})=0$ and $\mathrm{f}^{\prime \prime}(\mathrm{x}) < 0$ it is maximum
$\mathrm{f}^{\prime \prime}(\mathrm{x})=-2 < 0$
so, $\mathrm{f}^{\prime}(\mathrm{x})=0$ will give maximum value. so, $12-2 x=0 \Rightarrow x=6$ and $x+y=12 \Rightarrow y=6$
Hence, $\mathrm{y}=6$ and $\mathrm{f}(\mathrm{x})=12 \mathrm{x}-\mathrm{x}^{2}=12 \times 6-36=36$
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