Absolute minimum?

Question

y = x^3 - 15x^2 + 27x where -1<=x <= 2, now find the absolute minimum of this function.

Answer 1

y = x³ -15x² + 27x

y' = 3x² -30x +27

On a range, y has a minimum either at the extrema (x=-1 or x=2) or at a local minimum (y'=0).

Solve y'=0
=> 3x² -30x +27 = 0
=> x² -10x +9 = 0
(x-9)(x-1) = 0
x =1 or 9
x =9 is outside the range.
x=1 => y(1) = 1 -15 +27 = 13
(since y''<0, this is a point of inflection, not a minimum)

And testing y at the domain limits:
y(-1) = (-1)³ -15(-1)² + 27(-1) = -1 -15 -27 = -43
y(2) = (2)³ -15(2)² + 27(2) = 8 -60 +54 =2

Thus y has an absolute minimum of -43 (at x= -1).

Answer 2

Take the First Derivative...
Y'(x)= 3x^2 - 30x + 27
Then find where this is equal to zero. The values of x where it is equal to zero, indicate a possible maximum or minimum.
So....
0= 3x^2 - 30x + 27
All over three (makes it simpler to factor)

0= x^2 - 10 + 9

0= ( x - 9)(x - 1)
So x=9, x=1

We use these x values to find the critical point. We need the y coordinates so we plug them back into the original equation.

First I'll do 9.

Y=9^3 -15(9)^2 + 27(9) = -243.

Now 1.

Y= 1 - 15 + 27 = 13.

So our critical points are (9, -243) and (1,13)

We now have to find the second derivative to determine the concavity, which tells us if the point is a local max or local min. If the second derivative,Y''(x), is negative at the critical point, the point is a local max. If the second derivative is positive, the point is a local min.

Y''(x)= 6x -30
When x=9,
Y''(x)= 24, a positive. This is a local min.
When x=1,
Y''(x)= -24, a negative. This is a local max.

We have found one point that is a local min, (9, -243).
HOLD UP! We can't be positive quite yet......
Since the first derivative tells us points that are local max/mins IN RELATION TO THE X VALUES SURROUNDING THEM..... from this method we can not be certain that the endpoints aren't local mins/maxs.

Looking at the range for x, shows that 9 cannot be an answer, and that x=-1, and x=2 must be checked.
X=2,
Y= 2.
X= -1,
Y= -43.

This gives us the points (2, 2) and (-1, -43).
Therefore, the absolute minimum is (-1, -43)
Enjoy!

Answer 3

given y= x^3 - 15 x^2 +27x
to find minumum or maximum find y'

y'= 3x^2 -30x +27

put y'=0;

=> 3x^2 -30x +27 =0

=> x^2 -10 x+9=0 (dividing above eqn by 3)
=>(x-1)(x-9)=0
==> x=1 or x=9

since -1<=x<=2, we reject x=9;
we accept x=1
now find y'' at x=1

y''=6x-30
at x=1, y'' = 6(1) -30
==> y''= -24 which is less than zero
there fore we dont have absolute minimum value at x=1.

now check at boundary points i.e, -1 and 2

y(-1) = (-1)^3 - 15(-1)^2 +27(-1)

= -1-15-27
=-43 <0

y(2)= 8-60+54
= 2 >0

clearly y(-1)
therefore the given eqn will have absolute minimum value at y=-1, and the value is -43.

hope you got it.

Answer 4

To find the absolute minimum, you find any local extrema, then compare the values at those extrema, and the values at the endpoints to pick the lowest value.

Local extrema are where the derivative is zero.
y' = 3x^2 -30x + 27 0 = x^2 -10x +9
So there are extrema where x = 1 and where x = 9
9 is outside the range of interest, so we can ignore it.

Possible minimums are at -1, 1 or 2
y(-1) = -43
y(1) = 13
y(2) = 2

Absolute minimum is -43, occuring when x = -1

Answer 5

y' = 3x^2 - 30x + 27
3(x^2-10+9) = 0
3(x-9)(x-1) = 0
x = 1 and 9; but x = 1 is the only answer inside the domain

So the critical points of this are: 1, -1, and 2

1:
y'' = 6x - 30
6*1 - 30 = -24
Negative so this is a maximum.

So, we have a absolute minimum at x = -1