maxium and minium problem (1 Viewer)

[tradewind]

i don't get max and min problems. How does one approach these kind of questions. I'm struggling to do this one.

A stone is thrown upward so that , at time T seconds after throwing, the height H metres is given by H=300T-50T^2. Find the time at which the max height is reached and find the max height?

 

[hatty]

usually you derive the given equation.
in this case :
dH/dt = 300-100t
you then set the derivative = 0 (for stationary points)

300 = 100t
:. t = 3

then test if t = 3 is a maximum point.

do this by find the 2nd derivative, which is -100
-100 < 0
and when the 2nd derivative is < 0
it is concave DOWN
therefore, at t = 3, it is a maximum point.


simply sub t = 3 back into the original equation for the height
proven

 

[fatmuscle]

isn't this just differentiation???

 

[fatmuscle]

oops too late
other stuff:

H =-50T^2 + 300T

to find when height is 0 or the time it takes for the object to come back to it's starting point

0 = -50T^2 + 300T
0 = -50T(T - 6)

T = 0 or 6


this doesn't have gravity does it?

 

[hatty]

this isn't applications of calculus
its 'geometrical' applications of calculus

 

[tradewind]

thanks, it makes sense now.

 

[~*HSC 4 life*~]

Originally posted by tradewind
i don't get max and min problems. How does one approach these kind of questions.

The general way to approach these questions:
1. draw a diagram
2. using the info given, try and make and equation that connects two or more variables
3. solve for the variable
4. using FDT or SDT check if its a max/min

i can't seem to do this q either...2 exams today my brains gone a bit dead:

a rectangle is cut from a circular disc of radius 6cm. Find the area of the largest rectangle that can be produced

 

[CM_Tutor]

Originally posted by ~*HSC 4 life*~
The general way to approach these questions:
1. draw a diagram
2. using the info given, try and make and equation that connects two or more variables
3. solve for the variable
4. using FDT or SDT check if its a max/min

i can't seem to do this q either...2 exams today my brains gone a bit dead:

a rectangle is cut from a circular disc of radius 6cm. Find the area of the largest rectangle that can be produced

I would modify this method slightly, to:

1. Draw a diagram
2. Define variables (if needed) and set up problem
3. Find an expression, any expression, for whatever you seek to maximise or minimise
4. This expression will almost always have 3 variables - go back to the question, and identify the piece of information that you have not used. This should now be used to eliminate one variable.
5. Calculus - find dy/dx, and solve for dy/dx = 0 - before diving in to the problem, think. In many cases the calculus can be simplifed by noting things like the minimum value of y will occur at the same x value as the minimum value of y² - more on this below
6. Test nature to ensure it is a max / min as needed - If it isn't, check whether the problem has a restricted domain, and test the ends. For example, if you're asked for a max, and find only a min, then it's concave up, and so the max on the domain is at one of the ends - see, for example, the 1992 HSC, Q10.

I'll use the above example to illustrate later today.

 

[CM_Tutor]

Originally posted by ~*HSC 4 life*~
a rectangle is cut from a circular disc of radius 6cm. Find the area of the largest rectangle that can be produced

Step 1: Diagram
Mine is a circle, with a rectangle ABCD inscribed within it. The centre of the rectangle (ie. the intersection of the diagonals) is at the centre O. This is inevitably true provided A, B, C, and D are on the circumference (think about plane geometry if you're not convinced), and they must be on the circumference to be the largest possible rectangle.
Distances OA, OB, OC, and OD are marked as being equal to the radius, R = 6 cm, and the four vertices of the rectangle are marked as right angles.

Step 2: Define variables (if needed) and set up problem
Let the length of the rectangle, AB, be L cm, and its width, BC, be W cm. Let the area of the rectangle be A cm².
So, I marked AB = CD = L on the diagram, and AD = BC = W.
(Note: it should be obvious that L close to 2R in length leads to a small rectangle in area, as W is close to 0, as does L close to 0 with W close 2R - intuitively, symmetry demands that the largest rectangle will be the square
L = W.)

Step 3: Find an expression, any expression, for whatever you seek to maximise or minimise
Clearly, A = LW

Step 4: This expression will almost always have 3 variables - go back to the question, and identify the piece of information that you have not used. This should now be used to eliminate one variable.
The information that we have not yet used is the fact that the radius is R = 6 cm. Looking at the diagram, ABD is a right angled triangle, and applying Pythagoras' theorem we get L² + W² = (2R)² = 144. Making W the subject, this means that W = sqrt(144 - L²), as W > 0, and so:
A = L * sqrt(144 - L²)

Steps 5 and 6: Calculus and the answer
Firstly, the obvious way:

A = L * sqrt(144 - L²)
dA/dL = sqrt(144 - L²) * 1 + L * (1 / 2) * (-L) / sqrt(144 - L²), using the product and chain rules
= [(sqrt(144 - L²))² - L²] / sqrt(144 - L²)
= 2(72 - L²) / sqrt(144 - L²)

Put dA/dL = 0: 2(72 - L²) / sqrt(144 - L²) = 0
72 - L² = 0
L = sqrt(72), as L > 0
L = 6 * sqrt(2)

Nature - don't want to find d²A/dL², so use the sign test:
<table width=66% cellpadding=2>
<tr><td><font size=2>L<td width=15% align=center colspan=3><font size=2>6 * sqrt(2) - epsilon<td width=15% align=center colspan=3><font size=2>6 * sqrt(2)<td width=15% align=center colspan=3><td><font size=2>6 * sqrt(2) + epsilon
<tr><td><font size=2>dA/dL<td align=right><font size=2>(+)<td width=15%><td align=right><font size=2>0<td width=15%><td align=right><font size=2>(-)<td width=15%>
<tr><td><font size=2>Shape<td align=right><font size=2>/<td width=15%><td align=right><font size=2>^_<td width=15%><td align=right><font size=2>\<td width=15%>
</table>

So, we have a max at L = 6 * sqrt(2) cm
So, W = sqrt(144 - L²) = sqrt(144 - 72) = sqrt(72) = 6 * sqrt(2) cm
So, the rectangle of maximum area is the square with L = W = 6 * sqrt(2) cm, and A = (6 * sqrt(2))² = 72 cm².

Secondly, the quick way:

Note that the form of A would be much easier to differentiate if the sqrt were gone. Now, the maximum area A will occur at the same L value as the maximum value of A², so we can do calculus on A², instead...

A² = [L * sqrt(144 - L²)]² = 144L² - L⁴
dA²/dL = 144 * 2L - 4L³ = 4L(72 - L²)

Put dA²/dL = 0: 4L(72 - L²) = 0
L = sqrt(72), as L > 0
L = 6 * sqrt(2)

Nature - second derivative test is now fine:

d²A²/dL² = 144 * 2 - 4 * 3L² = 12(24 - L²)
At L = 6 * sqrt(2), d²A²/dL² = 12[24 - (6 * sqrt(2))²] < 0 ===> MAX

So, we have a max at L = 6 * sqrt(2) cm
So, W = sqrt(144 - L²) = sqrt(144 - 72) = sqrt(72) = 6 * sqrt(2) cm
So, the rectangle of maximum area is the square with L = W = 6 * sqrt(2) cm, and A = (6 * sqrt(2))² = 72 cm².

 

[~*HSC 4 life*~]

Thank you sooo much CM tutor!

your method is heap systematic, thanks for that..geez..is there anything you DON'T know? lol

thumbs up

 

[CM_Tutor]

Originally posted by ~*HSC 4 life*~
is there anything you DON'T know? lol

Well, if someone would tell me how to do tables in these posts, that'd be good. The one above is cut and pasted from a Lazarus post, and then I tried to modify it, but clearly I haven't got it right. Oh well...

Glad you like the method - I think the systematic approach is the key to most Maths problems - And yes, there are things I don't know, but I do know a fair bit about HSC Maths and Chemistry.

 

[CM_Tutor]

As a further comment on step 4, careful thought should be given when deciding which variable to eliminate. Of course, in many questions it is mandated by the question, so no thought is required. However, when it is not, much work can be saved by eliminating the 'correct' variable.

Consider the following fairly typical problem:

A piece of wire of length L m is to be used to form a sector of a circle. Find the area of the largest possible sector that can be so formed.

Setting up the problem with a sector of radius R m, an angle of @ radians, and an area of A m² it's easy to see that step 3 gives A = R²@ / 2.

Step 4, the unused piece of information, is that the perimeter is L m, and hence L = 2R + R@, but we have the choice of eliminating R or @.

Eliminating R gives A = L²@ / 2(@ + 2)², and you need to find dA/d@.

Eliminating @ gives A = R(L - 2R) / 2, and you need to find dA/dR.

Both will lead to the same result, R = (L / 4) m, @ = 2 rad, and A_max = (L² / 16) m². However, in my opinion, the calculus is much easier if you eliminate @.

Just something for you to think about...