Week 4 | Practice Assingment | Question 12

[viju...@gmail.com]

Could anyone explain Question 12 of Practice Assignment  of Week 4?

A farmer has a wire of length 576 metres. He uses it to fence his rectangular field to protect it from animals. If he fences his field with four rounds of wire, and
the field has the maximum area possible to accommodate such a fencing, what is the area (in square metres) of the field?

Thanks.

[Anand Iyer]

What did you try so far?

[viju...@gmail.com]

As per my solution the Area of Square equation is 72l - l^2 and Axis of Symmetry is 72/2 = 36 (......l). So in order to get the max Area...if I use value of 'l' ( = 36) in Area of Square equation (72l - l^2)...My answer differs from what is mentioned in solution document. Though Equation is same???

Regards,

Vijender

[Anand Iyer]

First of all, it's not a square.  So, the equation for finding area is different.

Secondly, why are you considering axis of symmetry?  what's the purpose?

Essentially, you need to derive a quadratic equation from whatever is given, and need to maximize the area from this.

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Cheers,

Anand

[viju...@gmail.com]

My mistake, it was a typo...I mean rectangle....and its the Quadratic Equation I mentioned (72l - l^2) in my post. And Axis of Symmetry should be used to determine...either minumum or maximum value of a quadratic function...isnt it?

On Friday, November 6, 2020 at 5:41:37 PM UTC+5:30, Anand Iyer wrote:

First of all, it's not a square.  So, the equation for finding area is different.

Secondly, why are you considering axis of symmetry?  what's the purpose?

Essentially, you need to derive a quadratic equation from whatever is given, and need to maximize the area from this.

On Fri, Nov 6, 2020 at 4:34 PM <viju...@gmail.com> wrote:

As per my solution the Area of Square equation is 72l - l^2 and Axis of Symmetry is 72/2 = 36 (......l). So in order to get the max Area...if I use value of 'l' ( = 36) in Area of Square equation (72l - l^2)...My answer differs from what is mentioned in solution document. Though Equation is same???

Regards,

Vijender

On Friday, November 6, 2020 at 9:03:57 AM UTC+5:30, Anand Iyer wrote:

What did you try so far?

On Thursday, November 5, 2020 at 10:37:23 PM UTC+5:30 viju...@gmail.com wrote:

Could anyone explain Question 12 of Practice Assignment  of Week 4?

A farmer has a wire of length 576 metres. He uses it to fence his rectangular field to protect it from animals. If he fences his field with four rounds of wire, and
the field has the maximum area possible to accommodate such a fencing, what is the area (in square metres) of the field?

Thanks.

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Cheers,

Anand

[Malabika Guha Mustafi]

@ Viju

We calculate vertex coordinate for maximum and minimum( the  equation of line considering the vertex x coordinate is axis of symmetry).

First compute the two sides relation from perimeter formula

Now , putting the two value in area equation so  you will get quadratic equation.

Compute vertex. (for maximum area)

Then Compute area.

Hope, it helps you.

[Anand Iyer]

I was about to type the steps out...and then realized it's already solved in practice assignment solution.  Did you check it out?

https://drive.google.com/file/d/1V3lT5SsDKZkPnbwTEW-ShlvZowZHVUpU

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Cheers,

Anand

[viju...@gmail.com]

Thank you so much both of you for your remarks....It was a silly mistake at my part 

If I apply value of x = 36 (Axis of Symmetry ) in equation 72l - l^2 (which I mentioned in my initial post)....I am getting the right answer....(I hate my typos...added instead of substracting in the end...and made a fool out of myself)

Pardon me and thanks again.

Regards,

Vijender

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Cheers,

Anand

[prasanth PS]

HI ,

how we are deriving the below for the same questions?

maximum area of the eld (Amax)= (-b^2 / 4*a) +C

As per my understanding its A(max) = -b/2A . Then how its changed to (-b^2 / 4*a) +C.

Please clarify...

Rgds

prasanth.

[Malabika Guha Mustafi]

@Prasant

We get , y= -  square(b)/4a  +  c ( for maximum value or minimum value after putting x=-b/2a) as follows:-

y= a(xsquare ) + bx+c

= a. square(-b/2a) +   b. (-b/2a) +c

=square(b)/4a  - square(b)/ 2a  + c

= -  square(b)/4a + c [ if you can remember , it is direct formula to get maximum/minimum]



[Anand Iyer]

It's not changed.  It's a different formula.  Note the -b/2a (previous one) finds the maximum x-value.  

Malabika, can you help?  I noticed you'd replied to this earlier...in another thread.

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Cheers,

Anand

[Malabika Guha Mustafi]

@ Anand, already replied :)

[viju...@gmail.com]

Hi Prasanth,

I could understand your question here. Its bit confusing the way its explained in the Solution Document by support team. However, if you stick to formula you mentioned (-b/2A), you should get Axis of Symmetry as 36. And your Quadratic Equation should be 72l - l^2, now if you apply value of x = 36, you should get the right answer.

Hope it helps.

Regards,

Vijender

[Malabika Guha Mustafi]

@Viju, it is explained. It is the shortest method.Just check the above answers. It may help during exam when time is constrained.

[viju...@gmail.com]

Indeed. Couldnt agree more.  

Regards,

Vijender

[Malabika Guha Mustafi]

Another short cut method,

for equation of a perpendicular line  passing through a point of a given line in general form.

let , the line is in Ax+ By + C=0 form

then , the perpendicular line equation through point(x0, y0) will be

y-y0 = B/A(x-x0) 

[ It is derived from point slope form, y-y0= m(x-x0)

 Here m will be -1/(slope of given line )since it is perpendicular of that line.

slope of the given line =-A/B

so, m= B/A.]

[Anand Iyer]

CAUTION: some bit of plain advice ahead...

While short-cut methods are good to know, there's bound to be a lot of confusion about what goes where, and what must be used when?

In my opinion, it's best to avoid short-cuts.  Work from first principles.  That's the safest, no confusions, and sometimes the simplest, 

If you ask me, this is the single biggest reason the majority of the school/college students in our country aren't able to think...and find math so tough.  

If it's tough, it's tough...get over it.  There are so many avenues nowadays that make learning all of this.  Once you get over it, once you crack the code, you'll find it much simpler, and even enjoyable...

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Cheers,

Anand

[Malabika Guha Mustafi]

@Anand

That's the reason I posted here with explanation not in separate thread. Who is other member of the thread I am quite sure they can understand the first principle to solve the problem.

[Anand Iyer]

Got it Malabika.

However, this post is not restricted to only to few of us, everyone can read and follow this...

I hope you didn't take my reply as an offence...

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Cheers,

Anand