Maximum / minimum question (1 Viewer)

enigma_1

~~~~ Miss Cricket ~~~~
Joined
Feb 27, 2013
Messages
4,281
Location
Lords
Gender
Female
HSC
2014
Math question.png

How do I find the expression for this?

Thanks
 

ChillTime

Member
Joined
Dec 9, 2013
Messages
40
Gender
Male
HSC
2005
Basically you want to subtract the 2 curves:

AB = y1 - y2
= [(x-3)^2+7]-[x(4-x)]
= 2x^2 -10x +16

(Length now varies as a function of x)

Next you want to differentiate

Let h = AB = 2x^2 -10x +16

dh/dx = 4x - 10

Stationary points, i.e. gradient is 0

dh/dx = 0
4x - 10 = 0
x = 2.5

Find nature of stationary point - take 2nd derivative

d^2h/dx^2 = 4 < 0, therefore the curve is concave up, and the curve has a minimum value. [Note, if this was 4x you would need to sub in x=2.5]

Minimum length of AB:

AB_min = h(2.5) = 2(2.5)^2 - 10(2.5) + 16 = 3.5 units
 

HeroicPandas

Heroic!
Joined
Mar 8, 2012
Messages
1,547
Gender
Male
HSC
2013
u subtracted top curve from bottom which is correct but some other people might ask why don't u subtract bottom curve from top curve?

by looking at the curve and points A and B, u can say that the y-coord of A is greater than the y-coord of B. so if u subtract bottom curve from top u get a negative value (because SMALL minus BIG = super SMALL, e.g. 1 - 8 = -7)

y_A > y_B

y_B < y_A

y_B - y_A < 0

AB < 0 which is weird...distance cannot be negative
 

bottleofyarn

Member
Joined
Sep 22, 2013
Messages
50
Gender
Male
HSC
2013
Use a subtraction of the y coordinates ie y1 - y2 to find the length, since the x value is the same. Getting a quadratic from this, you can manipulate and graph instead of differentiating. My answer is 7/2.
 
Last edited:

ChillTime

Member
Joined
Dec 9, 2013
Messages
40
Gender
Male
HSC
2005
Use a subtraction of the y coordinates ie y1 - y2 to find the length, since the x value is the same. Getting a quadratic from this, you can manipulate and graph instead of differentiating. My answer is 7/2.
Indeed you can and it's shorter, but if this is the min/max topic, the featured method is differentiation.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top