Stuck on a maths question? Help please?

Question

Find the coordinates of the maximum and minimum points on the curve [y=x^2(5-x)^3, distinguishing between them.

^2(squared) and ^3 (cubed)

There are actually 2 points (x,y) and (x,y). You have to be really clever at maths to do this.

You have to use the Product Rule and differentiate

Answer 1

Using Calculus, a function's maxima or minima or point of inflexion may be found by setting the first derivative = 0. If the second derivative < 0, we will have a maxima. If the second derivative is > 0 we will have a minima. If the second derivative = 0, we have a point of inflexion.

So, using the chain rule:

dy/dx = 2x(5-x)^3 - 3x(5-x)^2 = 5x(5-x)^2(2-x)

Setting dy/dx = 0 yields:

5x(5-x)^2(2-x) = 0

Now we can see that this equation has the solutions
x = 0, x = 5 or x = 2. So using the original equation we can see that the points (0, 0), (5, 0) and (2,108) are all maxima or minima or points of inflexion.

We now find the second derivative again using the chain rule:

d^2y/dx^2 = 5(2-x)(5-x)^2 - 5x(5-x)^2 - 10x(2-x)(5-x)

Lets consider the first point, (0,0). Substitution of x=0 into the second derivative gives d^2y/dx^2 = 250. A positive number, so the point (0,0) is a minima of the function.

The second point, (5,0) yields d^2y/dx^2 = 0. This is a point of inflexion.

The third point, (2,108) yields d^2y/dx^2 = -90. A negative number, hence this point is a maxima.

Hope that helps!

Answer 2

use dy upon dx by using the UV rule...equal dy upon dx to zero.if am not wrong,you`ll obtain 2 answers for x.then replace in original equation to get value of y.for maximum,the answer must include the minus sign and a plus sign for the minimum point...hope i`ve been a bit helpful to you dear...or else use the method of completing the square which is more simple.

Answer 3

y = x^2(5-x)^3

y' = 2x(5-x)^3 + x^2[-3(5-x)^2]
y' = 2x(5-x)^3 - 3x^2(5-x)^2
y' = x(5-x)^2 [2(5-x) - 3x]
y' = x(5-x)^2 (10-5x)

y'' = (5-x)^2 (10-5x) + 2(5-x)(-1)x(10-5x) + x(5-x)^2(-5)
y'' = (10-5x)(5-x)^2 - 2(5-x)(10-5x) - 5x(5-x)^2
y'' = 2(5-x)^3 - 4(5-x)^2 - 5x(5-x)^2

At the max and min points, the gradient = 0, so y'=0.
x(5-x)^2 (10-5x) = 0
x = 0 or 5 or 2

When x=0, y=0, y''=150 > 0 => minimum point
When x=5, y=0, y''=0 => point of inflection
When x=2, y=108, y''=-72 => maximum point

Minimum point: (0,0)
Maximum point: (2, 108)

Answer 4

rewrite x^2(5-x)^3 as 125x^2-75x^3+15x^4-x^5

1st diff 250x-225x^2+60x^3-5x^4=0 at the turning points

three different solutions are x=0,2 or 5 at ist diff=0 at the turning points

2nd diff 250-450x+ 180x^2-20x^3

at x =0, 2nd diff =250,hence min

at x =2, 2nd diff=-90,hence max

at x =5, 2nd diff=0,hence point of inflection

(0,0) is the min point,(2,108) is the max point,(5,0)is the point of inflection

Answer 5

Are you sure you have this equation correct? If there is to be one max and one min point it should be something like y = x cubed + x squared + .........

Answer 6

is like the questionis not correct,re-check again. the curve is endless

Answer 7

the curve cannot be perfectly defined n yeah.........
looks like d curve extends 2 infinity......

Answer 8

Yeah, what antoine said, he beat me to it

Answer 9

Do you own bloody home work

Answer 10

(0,0) min and (3,108) max