Question

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

Then the minimum value of $y$ in this period-

• A. $-\infty$
• B. $276$
• C. $-124$
• D. $-128$

Answer 1

The correct option is D $-128$

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 minimum value of the function occurs at $x=6$. So the value of the function ar this point is,

$\Rightarrow y\left(6\right)={2.6}^3-{21.6}^2+36.6-20$

$\Rightarrow y\left(6\right)=432-756+216-20$

$\Rightarrow y\left(6\right)=648-776$

$\Rightarrow y\left(6\right)=-128$ .....Answer

Thus the minimum value of the function is -128.