Applications of Maximization and Minimization (1 Viewer)

[xXnukerrrXx]

I am having trouble with a question in the maximization and minimization questions using calculus. It comes from Question 12 and 13 of year 11 Cambridge
Q 12) A window consisting of 6 equal rectangles
i) Let entire frame have h height and width y meters. If 12 meters of frame is available show that y=1/4(12-3h)
ii) Show area is A=3h-(3/4*h^2)
Q 13) Prism has volume of 4374m^3 find dimensions of frame so that minimal amount of steel is used. GIVEN: length: 3x Width: x and Height is h

Thanks for your help!!!

 

[Drongoski]

For Q12: main difficulty is not the maths - but in trying to understand what the question is saying; I'm still confused myself on 1st reading. The maths, once you understand the question, is probably trivial.

 

[xXnukerrrXx]

thats the same problem im having :/

 

[nightweaver066]

Unsure on question 12. Here's the question from the Cambridge book though, gives more detail

Question 13 though, you know the volume is

Since that is the volume and we know the volume is all the dimensions multiplied with each other,

So we can say:


Here is our first equation.

Now the other equation, we're going to use the perimeter of the shape (since it has a length, width and height, assuming it's a rectangular prism of some sort?)



You should know what to do next and hopefully my assumption was right..

If you don't know, just PM me or post here or something.

 

[xXnukerrrXx]

how do you do it?

 

[nightweaver066]

Okay well with our two equations, lets rearrange the first one.


Dividing both sides by 3x^2


Subbing this in to the second equation,




Now differentiating,



Second derivative,



For minimum/maximum perimeter,






Subbing in to second derivative to show whether it is maximum/minimum.


Subbing to the first equation,


The dimensions of the rectangular prism are for minimum amount of steel to be used.

 

[Drongoski]

how do you do it?

Q12 now becomes easy. You have omitted vital info - the diagram. Quite annoying really. Had I been given the diagram I'd been able to show you straightaway. Be careful how you re-present a question in foture.

Q12

(a) total length of frame = 4y + 3h = 12

.: y = 1/4 * (12-3h)

(b) Area = A(h) = length x breadth = h * y = y * h = 1/4 * (12-3h) * h = 3h - 3/4 * h²

(c) For maxinum A(h) dA/dh = 3 - 3/2 * h = 0 ==> h = 2

.: dimensions of frame, viz h & y are: 2 and 1.5 (if, in my haste, I've not made a mistake)

 

[maths lover]

just out of curiosity can these questions be done without calculus, because we have learnt the topic but not calculus.

 

[nightweaver066]

Well to solve minima and maxima questions, you need to find the maximum and/or minimum point of an equation. In quadratics, there is only one maximum/minimum point and they can easily be found because it is the vertex. I'm guessing the only type of questions solvable without calculus are quadratics or equations reducible to quadratics.

These types of questions however have much more complex graphs where it would be difficult to determine the maximum/minimum points in the graph without the use of calculus (in order to find the maximum/minimum points).

 

[SpiralFlex]

You guys post too fast. Also the trickiest bit to these questions are not how to do it. But deriving and understanding the formula. If you can connect two things together. They will "click."

Q12 now becomes easy. You have omitted vital info - the diagram. Quite annoying really. Had I been given the diagram I'd been able to show you straightaway. Be careful how you re-present a question in foture.

Q12

(a) total length of frame = 4y + 3h = 12

.: y = 1/4 * (12-3h)

(b) Area = A(h) = length x breadth = h * y = y * h = 1/4 * (12-3h) * h = 3h - 3/4 * h²

(c) For maxinum A(h) dA/dh = 3 - 3/2 * h = 0 ==> h = 2

.: dimensions of frame, viz h & y are: 2 and 1.5 (if, in my haste, I've not made a mistake)

Test for maximum?

 

[Drongoski]

Strictly better to show it'a a maximum. Since A(h) is a quadratic in h with negative coeff in h ² the extremal point is of course a max point.

 

[xXnukerrrXx]

calculus is easy once you get the hang of it, the hard bit is to interpret it properly. Thanks for the help guys! i managed to find it after coming back to it 10 mins ago (very annoyed seeing how easy it actually was). Still thankyou very muchhh