Question 39

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Jambmaths question: 

Determine the maximum value of $y=3{{x}^{2}}-{{x}^{3}}$

Option A: 

0

Option B: 

2

Option C: 

4

Option D: 

6

Jamb Maths Solution: 

$\begin{align}  & y=3{{x}^{2}}-{{x}^{3}} \\ & \frac{dy}{dx}=6x-3{{x}^{2}} \\ & \text{At stationary point }\frac{dy}{dx}=0 \\ & 6x-3{{x}^{2}}=0 \\ & 3x(2-x)=0 \\ & x=0,\text{ }x=2 \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{d}{dx}(6x-3{{x}^{2}})=6-6x \\ & \text{At }x=0,\text{  }\frac{{{d}^{2}}y}{d{{x}^{2}}}=6-6(0)=6>0\text{   }\!\!\{\!\!\text{ min point }\!\!\}\!\!\text{ } \\ & \text{At }x=2,\text{   }\frac{{{d}^{2}}y}{d{{x}^{2}}}=6-6(2)=-6<0\text{   }\!\!\{\!\!\text{ max point }\!\!\}\!\!\text{ } \\ & \text{At }x=2\text{ will give the maximum value of }y \\ & \therefore {{y}_{\max }}=3{{(2)}^{2}}-{{(2)}^{3}}=12-8=4 \\\end{align}$

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