Question 38

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Jambmaths question: 

What is the value of x will make the function $x(4-x)$ a maximum?

Jamb Maths Solution: 

$\begin{align}  & y=x(4-x) \\ & y=4x-{{x}^{2}} \\ & \frac{dy}{dx}=4-2x \\ & \text{At turning point }\frac{dy}{dx}=0 \\ & 4-2x=0 \\ & x=2 \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{d}{dx}\left( \frac{dy}{dx} \right)=\frac{d}{dx}(4-2x)=-2 \\ & \text{Since }\frac{{{d}^{2}}y}{d{{x}^{2}}}<0,\text{ so at }x=2\text{ will make the function a} \\ & \text{maximum}\text{.} \\\end{align}$

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