VCE Methods Question Thread!

[Kel9901]

Hi, guys! I have a SAC soon on Calculus and I'm having trouble understanding a specific type of questions on optimisation problems.
I;m pretty okay at most optimisation problems but questions that go "Find the area of the largest "so and so" that can fit inside "so and so" or "Find the dimensions of a largest triangle that can be inscribed in "so an so" " and such make me go crazy

I pulled out a couple of questions that I don't know and was wondering if anybody would be so kind to go through them, step by step

1. Find the area of the largest rectangle that fit inside the a right angled triangle with height 3cm and base 4 cm (Ans: 3cm squared)

2. Find the dimensions of the isosceles triangle of largest area that can be inscribed in a circle of radius r. (answer: base: sqr(3), height 3r/2)

3. A right circular cylinder is inscribed in a cone with height h and base radius . Find the largest possible volume of such a
cylinder.

I would really appreciate it if someone could answer!

1. Right angled triangle with the right angle at the bottom left:
Let l be the length from the bottom left point, and h be the height.
A=lh
At the top of the triangle, l=0, h=3 ie (0,3)
At the bottom-right of the triangle, l=4, h=0 ie (4, 0)
The equation relating l and h is h=-3/4 l +3
Subbing into the area formula:
A=l(-3/4 l +3)=-3/4 l²+3l

dA/dl=-3/2 l+3=0
3=3/2 l
l=2
At l=1.9 dA/dl=0.15>0 and at l=2.1 dA/dl=-0.15<0 hence the SP at l=2 is a maximum
A=-3/4 l²+3l=-3/4 (4)+3(2)=3 cm²

2.
http://puu.sh/irMgF/2c3be6164b.png hope you can understand my poorly drawn diagram
Area of the isosceles triangle is A=1/2 (2x)y=xy where 2x is the base and y is the height

From the diagram, (y-r)²+x²=r²
x=sqrt(r²-(y-r)²)=sqrt(r²-y²+2yr-r²)=sqrt(2yr-y²) (as x>0)

A=xy=sqrt(2yr-y²)y
dA/dy=(2r-2y)y/(2sqrt(2yr-y²))+sqrt(2yr-y²)=0 (product + chain rule)
multiplying both sides by sqrt(2yr-y²):
(r-y)y+(2yr-y²)=0
yr-y²+2yr-y²=0
3yr-2y²=0
3r-2y=0 as y>0
y=3r/2

x=sqrt(2yr-y²)=sqrt(3r²-9r²/4)=sqrt(3r²/4)=sqrt(3)r/2

so base=2x=sqrt(3)r and height=y=3r/2

3. You can pretty much think of it as inscribing a rectangle (cylinder) into a triangle (cone).

y=height of cylinder, x=radius of cylinder
at the top of the cone, x=0 and y=h ie (0, h)
at the bottom of the cone x=r and y=0 ie (r, 0)
Relationship is y=-h/r x + h

V=pi x²y=pi x² (-h/r x+h)=-pi h/r x³ + pi h x²

dV/dx=-3pi h/r x² + 2 pi hx=0
-3/r x+2=0 as x>0
x=2r/3

now sub into V formula
V=-8pi/27 hr² + 4pi/9 hr²=4pi hr²/27

[kinslayer]

In some sense, it's the 'real-world' information provided along with the function that determines the units.

Great post. To add another example to the above, the integral only represents a literal, physical area if f(x) and dx both have units of length

edit: Considering the average value of the function f(x) over the integral (a,b):



The integral is a continuous sum of quantities of the form , which has the units of f multiplied by the units of x. This is then divided by another quantity, b-a, which has the same units as x, if it has any.

Thus, A must have the same units as the function f(x) (if any).

[I_I]

1. Right angled triangle with the right angle at the bottom left:
Let l be the length from the bottom left point, and h be the height.
A=lh
At the top of the triangle, l=0, h=3 ie (0,3)
At the bottom-right of the triangle, l=4, h=0 ie (4, 0)
The equation relating l and h is h=-3/4 l +3
Subbing into the area formula:
A=l(-3/4 l +3)=-3/4 l²+3l

dA/dl=-3/2 l+3=0
3=3/2 l
l=2
At l=1.9 dA/dl=0.15>0 and at l=2.1 dA/dl=-0.15<0 hence the SP at l=2 is a maximum
A=-3/4 l²+3l=-3/4 (4)+3(2)=3 cm²

2.
http://puu.sh/irMgF/2c3be6164b.png hope you can understand my poorly drawn diagram
Area of the isosceles triangle is A=1/2 (2x)y=xy where 2x is the base and y is the height

From the diagram, (y-r)²+x²=r²
x=sqrt(r²-(y-r)²)=sqrt(r²-y²+2yr-r²)=sqrt(2yr-y²) (as x>0)

A=xy=sqrt(2yr-y²)y
dA/dy=(2r-2y)y/(2sqrt(2yr-y²))+sqrt(2yr-y²)=0 (product + chain rule)
multiplying both sides by sqrt(2yr-y²):
(r-y)y+(2yr-y²)=0
yr-y²+2yr-y²=0
3yr-2y²=0
3r-2y=0 as y>0
y=3r/2

x=sqrt(2yr-y²)=sqrt(3r²-9r²/4)=sqrt(3r²/4)=sqrt(3)r/2

so base=2x=sqrt(3)r and height=y=3r/2

3. You can pretty much think of it as inscribing a rectangle (cylinder) into a triangle (cone).

y=height of cylinder, x=radius of cylinder
at the top of the cone, x=0 and y=h ie (0, h)
at the bottom of the cone x=r and y=0 ie (r, 0)
Relationship is y=-h/r x + h

V=pi x²y=pi x² (-h/r x+h)=-pi h/r x³ + pi h x²

dV/dx=-3pi h/r x² + 2 pi hx=0
-3/r x+2=0 as x>0
x=2r/3

now sub into V formula
V=-8pi/27 hr² + 4pi/9 hr²=4pi hr²/27

Thank you, Kel9901!!!~!!

I get it! I GET IT I never thought I would. Thanks for the diagram- It's pretty tricky to draw on the computer

[Apink!]

Hi

Could someone please help me with this question?

I don't understand why the answer is A.. I thought there would be a stationary point of inflexion!

[kinslayer]

Hi

Could someone please help me with this question?

I don't understand why the answer is A.. I thought there would be a stationary point of inflexion!

Let F(x) be an antiderivative of f(x). Then its gradient is given by the derivative F'(x) = f(x). In other words, the gradient of the antiderivative of f(x) IS f(x). f(x) is positive between -a and a, hence answer A.

[Apink!]

Hi!

How would you know that f'(x) = f(x)? Is it how the question was worded or something?

Thank you for having patience with me

[kinslayer]

Hi!

How would you know that f'(x) = f(x)? Is it how the question was worded or something?

Thank you for having patience with me

That's just a notation thing. Often the antiderivative of a function f(x) is written with a capital F like F(x).

We don't need to know what the antiderivative of f(x) actually looks like, we just need to remember that the derivative of an antiderivative of a function is always the function itself.

All of the possible solutions are statements about the antiderivative's derivative, which is just f(x).

B, C, D and E are all incorrect because these imply that f(x) is zero at some point, but f(x) is non-zero between -a and a.

[I_I]

Hi guys, thought I gained what it takes to do optimisation problems with the help of Kel9901, but turns out it was my inflated hope

The questions I can't currently solve after thinking for like 20 minutes are:

A right circular cylinder is placed inside a sphere of radius 5cm. Find the largest possible volume of the cylinder

and

A right circular cylinder is placed inside a sphere of radius 5cm. Find the largest possible surface area of the cylinder

Thanks everyone

[knightrider]

I think that the average value of a function (by default) is unitless, because values of a function are unitless, and taking the average of these values shouldn't require putting some units in front of it. For example, if I asked you what the average of 1, 2, 3, 4 and 5 is, the answer would be 3, not 3 units.
Another way to look at it is that the average value of a function in the interval [a, b] is the value of the constant function which when integrated over [a,b] gives you the same result as integrating the original function over [a, b]. This is to show that the average value is just a number - you wouldn't write f(x) = 3 cm^2; you would write f(x) = 3.

In some sense, it's the 'real-world' information provided along with the function that determines the units. Consider a function, f(x) = x^3, dom(x) = (0, inf). While I can use this function to calculate the volume of a cube of side length x, that doesn't really mean that f(2) = 8 units^3. This is because the function f(x) = x^3, dom(x) = (0, inf) doesn't really have an intrinsic real-world meaning until we place some on it (by noticing that this function would be good at letting us calculate volumes of cubes, for example). Even though many functions have been inspired by real-world phenomena, at the end of the day, the functions are not intrinsically linked to them - they can be used as context-free mathematical objects, and hence do not require units. As another example, the average velocity of a car that uniformly accelerates at 1m/s^2 from 0m/s to 10m/s will be 5m/s over the first 10 seconds. But the average value of y = x over [0, 10] is just 5.

I suppose as a final comment, consider a (single-variable) function that maps from the real numbers to the real numbers (f: R -> R, f(x) = blah). What this means is that the function takes in a real number within its domain (this is the point of the first R), and spits out another real number (this is the point of the second R). What I'm trying to get at is that the function just spits out numbers without any units. Hence, the values of functions are unitless, and hence the average value of a function is unitless too.

Great post. To add another example to the above, the integral only represents a literal, physical area if f(x) and dx both have units of length

edit: Considering the average value of the function f(x) over the integral (a,b):



The integral is a continuous sum of quantities of the form , which has the units of f multiplied by the units of x. This is then divided by another quantity, b-a, which has the same units as x, if it has any.

Thus, A must have the same units as the function f(x) (if any).

Thanks so much nerdgasm  and kinslayer 

[Apink!]

That's just a notation thing. Often the antiderivative of a function f(x) is written with a capital F like F(x).

We don't need to know what the antiderivative of f(x) actually looks like, we just need to remember that the derivative of an antiderivative of a function is always the function itself.

All of the possible solutions are statements about the antiderivative's derivative, which is just f(x).

B, C, D and E are all incorrect because these imply that f(x) is zero at some point, but f(x) is non-zero between -a and a.

AHA!!!

*lightbulb moment*

Thank you so much kinslayer for helping me! (not for just this question, but for all the questions you answered (:  )

I get it now, thanks!!

[cosine]

f(x) = x^2+px+q

Find the coefficients p and q, if at x=2, the tangent is parallel to the x-axis.

From this information, I was able to work out p=-4, by deriving the equation, giving me the gradient function, and if the tangent is parallel to the x-axis, then the gradient is zero, so worked out p from there. But have no idea how to work out q?

[StupidProdigy]

Could you tell me the answer so I can check mine and show you how I got it if it is right?

[cosine]

Could you tell me the answer so I can check mine and show you how I got it if it is right?

I have no idea what the answer is, I lost the file for the worked solutions (Neap practice exam)

Can you show working out anyways?

[StupidProdigy]

I have no idea what the answer is, I lost the file for the worked solutions (Neap practice exam)

Can you show working out anyways?

Attached my working, really not certain in my answer and it would be good if someone could confirm the actual way to find it (not faithful in my thought  )

[cosine]

Attached my working, really not certain in my answer and it would be good if someone could confirm the actual way to find it (not faithful in my thought  )

How did you get equation of tangent to equal q-4??

Also, I graphed it with x^2+4x+4, unfortunately, that's not the correct answer

Bloody neap, so hard