Question

Let $y=2x^3-21x^2+36x-20$ and $(x\in (0,8\left)\right)$,

Then the maximum value of $y$ in this period is:

• A. $21$
• B. $-3$
• C. $-21$
• D. $3$

Answer 1

The correct option is B $-3$

The given information is:

$y=2x^3-21x^2+36x-20$ and x\in (0,8)

Differentiating the function w.r.t.$x$ and equating it to 0 to find the extremum points.

$\Rightarrow y^{\prime }=6x^2-42x+36=0$

$\Rightarrow 6x^2-42x+36=0$

$\Rightarrow x^2-7x+6=0$

$\Rightarrow (x-1)(x-6)=0$

So the two extremum points are $x=1$ and $x=6$.

To find whether the function has the maximum or minimum value at the given extremum point we find the double derivative of the function. If its is greater than zero at the extremum point then its a minimum point else a maximum point. The double derivative test fails if it is equal to zero.

So find the double derivative by differentiating $y^{\prime }$ we get,

$y^{\prime \prime }=12x-42$

$\Rightarrow y^{\prime \prime }\left(1\right)=12-42=-30$

$\Rightarrow y^{\prime \prime }\left(1\right)<0$. This means that maxima occurs at the point x =1

$\Rightarrow y^{\prime \prime }\left(6\right)=72-42=30$

$\Rightarrow y^{\prime \prime }\left(6\right)>0$.. This means that minima occurs at the point x=6.

The maximum value of the function occurs at $x=1$ which is equal to ,

$\Rightarrow y\left(1\right)=2{.1}^3-{21.1}^2+36.1-20$

$\Rightarrow y\left(1\right)=2-21+36-20$

$\Rightarrow y\left(1\right)=37-41$

$\Rightarrow y\left(1\right)=-3$ .....Answer

Thus the maximum value of the function is -3