# 3 questions

• Feb 23rd 2007, 09:24 PM
Nick744
3 questions
Hey guys,

Got some challange questions in my book, and they're challanging (for me atleast).

1. Sketch the curve y = x(x-2)^3 , showing any stationary points and inflexions.

2. Find the greatest and least values of the function f(x) = 4x^3 - 3x^2 - 18x in the domain -2 <= x <= 3.

3. Two circles have radii r and s such that r + s = 25. Show that the sum of areas of the circles is least when r = s.

• Feb 23rd 2007, 11:01 PM
CaptainBlack
Quote:

Originally Posted by Nick744
Hey guys,

Got some challange questions in my book, and they're challanging (for me atleast).

1. Sketch the curve y = x(x-2)^3 , showing any stationary points and inflexions.

first this has zeros a x=0, and x=2, the latter of multiplicity 3, so we know
that the curve looks like (x-2)^3 close tp x=2, and so we expect this point
to be a point of inflection.

Also as x goes to +/- infinity y goes to +infinity.

Now lets look for the stationary points:

dy/dx= (x-2)^3 + 3 x (x-2)^2 = (x-2)^2 [(x-2) + 3x]=2 (x-1) (2x -1)

Which is equal to zero when x=1/2, and when x=2. Then we can deduce
from shape of the cureve and the position of the zeros of y, that x=1/2 is a
minimum, and x=2 a point of inflection (we could confirm this with the
second derivative test if we wanted).

y~=-1.68, at the minima corresponding to x=1/2.

Putting this all together we get something like the attachment.

RonL
• Feb 23rd 2007, 11:08 PM
CaptainBlack
Quote:

Originally Posted by Nick744
2. Find the greatest and least values of the function f(x) = 4x^3 - 3x^2 - 18x in the domain -2 <= x <= 3.

A function differentiable function f(x) has a global maximum (minimum) on a closed interval
either at an end point of the interval, or as a calculus type local maximum in the interval.

So here

1. Compute the f(-2) and f(3)

2. Differentiate f(x) to get f'(x), solve f'(x)=0 to find the stationary points. Let the stationary points in [-2, 3] be x1, .. xn.

3. Compute f(x1), .. , f(xn), then the global maximum is:

max[ f(-2), f(3), f(x1), .. , f(xn) ].

The same procedure also finds the global minimum if max is rellaced by min in the last line above.

RonL
• Feb 23rd 2007, 11:20 PM
CaptainBlack
Quote:

Originally Posted by Nick744
3. Two circles have radii r and s such that r + s = 25. Show that the sum of areas of the circles is least when r = s.

Assume the circles do not overlap, then the sum of the areas is:

A=pi r^2 + pi s^2,

ans as r + s =25, we may substitute s = 25-r for s in the equation for A:

A = pi r^2 + pi (25-r)^2 = pi [ 2 r^2 -50 r +225 ]

To find the stationary points of A we differentiate wrt r to get A'(r), set this to zero to find the critical points:

A'(r) = pi [4 r - 50 ]

So A'(r)=0, has solution r=25/2. To show that this is a minimum we differentiate again to get A''(r)= 4 pi, which is >0, so r=25/2 is a minimum.

As r+s=25, when r=25/2 s=25/2, so we have a minimum area sum when r=s.

RonL