Area & Perimeter question: A fence is 80 m long and is used to enclose an area with five sides.

[bluefrog]

I have another question. This is not homework. I am TA, attempting to learn A level maths, so please bear with me.

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5 A fence is 80 m long and is used to enclose an area with five sides, as shown in the diagram:



Some of the dimensions of the fence are also shown. The area enclosed by the fence is [imath]\footnotesize{A\,\textrm{m}^2}[/imath].

a Find an expression for [imath]\footnotesize{y}[/imath] in terms of [imath]\footnotesize{x}[/imath].

b Prove that [imath]\footnotesize{A = 80x - x^2\,\left(1 + 2\sqrt{2\;}\right)}[/imath].

c Given that [imath]\footnotesize{x}[/imath] can vary, use calculus to find the exact maximum value of [imath]\footnotesize{A}[/imath]

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The answer for a) is shown in the back of the book as 5 a [imath]\;y = 4 - x\left(1 + \sqrt{2\;}\right)[/imath], but I don't get the same.

Pythagoras's theorem is [imath]a^2+b^2=c^2[/imath], so hence we must have [imath]x^2+x^2=(x+x)^2, \text{ and therefore } \sqrt{(x+x)^2} = 2x, \text{but I think this might be incorrect, since }, 80=2y+4x , \text{ is not what is required.}[/imath]

The only way for the answer to be satisfied would if [imath]80=2y+2x+2\sqrt{2x^2}, \text{ but I do not get how the third term on the RHS is obtained.}[/imath]

Any suggestions would be appreciated.

 

[The Highlander]

Pythagoras's theorem is [imath]a^2+b^2=c^2[/imath], so hence we must have [imath]x^2+x^2=(x+x)^2, \text{ and therefore } \sqrt{(x+x)^2} = 2x, \text{but I think this might be incorrect, since }, 80=2y+4x , \text{ is not what is required.}[/imath]

The only way for the answer to be satisfied would if [imath]80=2y+2x+2\sqrt{2x^2}, \text{ but I do not get how the third term on the RHS is obtained.}[/imath]

\(\displaystyle y = 40 - x(1+\sqrt{2})\) is perfectly correct,

Start with: \(\displaystyle 2y+2x+2\sqrt{2x^2}=80\) and re-arrange it (correctly) and you will get the book's answer. (I did.)

\(\displaystyle x^2+x^2\ne(x+x)^2\qquad(x+x)^2=4x^2~ \textbf{ not }~2x^2\)!!!

Hint: \(\displaystyle \sqrt{2x^2}=x\sqrt{2}\qquad\) ?

You get \(\displaystyle 2\sqrt{2x^2}\) on the RHS if you use Pythagoras properly!

The two shorter sides of the R-A Triangle are x so squaring them (and adding) gives x² + x² = 2x² and taking the square root of that gives \(\displaystyle \sqrt{2x^2}\) as the length of the hypotenuse (of which there are two on the perimeter), hence \(\displaystyle 2\sqrt{2x^2}\).    ?‍

 

[Steven G]

You should just know, after deriving it once, that there is a right triangle with sides 1, 1 and sqrt(2). Multiplying ALL sides by x (because you have 2 sides with x) you obtain a right triangle with sides x, x , xsqrt(2).
There should be no need to use the Phthagoras' theorem at all.

There are a few right triangles that you should just know. They are---
1,1, sqrt(2); 1, sqrt(3), 2 and 5, 12, 13

 

[BigBeachBanana]

I have another question. This is not homework. I am TA, attempting to learn A level maths, so please bear with me.

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5 A fence is 80 m long and is used to enclose an area with five sides, as shown in the diagram:

View attachment 36121

Some of the dimensions of the fence are also shown. The area enclosed by the fence is [imath]\footnotesize{A\,\textrm{m}^2}[/imath].

a Find an expression for [imath]\footnotesize{y}[/imath] in terms of [imath]\footnotesize{x}[/imath].

b Prove that [imath]\footnotesize{A = 80x - x^2\,\left(1 + 2\sqrt{2\;}\right)}[/imath].

c Given that [imath]\footnotesize{x}[/imath] can vary, use calculus to find the exact maximum value of [imath]\footnotesize{A}[/imath]

---

The answer for a) is shown in the back of the book as 5 a [imath]\;y = 4 - x\left(1 + \sqrt{2\;}\right)[/imath], but I don't get the same.

Pythagoras's theorem is [imath]a^2+b^2=c^2[/imath], so hence we must have [imath]x^2+x^2=(x+x)^2, \text{ and therefore } \sqrt{(x+x)^2} = 2x, \text{but I think this might be incorrect, since }, 80=2y+4x , \text{ is not what is required.}[/imath]

The only way for the answer to be satisfied would if [imath]80=2y+2x+2\sqrt{2x^2}, \text{ but I do not get how the third term on the RHS is obtained.}[/imath]

Any suggestions would be appreciated.

[imath]x^2+x^2 \neq (x+x)^2[/imath]

 

[The Highlander]

[imath]x^2+x^2 \neq (x+x)^2[/imath]

Hmmm, see Post #2. ?

 

[Steven G]

Hmmm, see Post #2. ?

How did you get the link to go to post #2????

 

[blamocur]

How did you get the link to go to post #2????

(Apologies for a poor formatting ): Use "share" icon , then copy the link at the bottom of the pop-up :

 

[The Highlander]

How did you get the link to go to post #2????

Simples!
1. Right-click on the Post Number
and
2. Choose: "Copy link address". Then use that (copied) URL for the link in your own post. ?‍

(Or use @blamocur's method. ?)​

 

[The Highlander]

How did you get the link to go to post #2????

Use "share" icon , then copy the link at the bottom of the pop-up :

But it's even better if you can link straight to the specific part (perhaps in the middle of a lengthy post) that you want to draw attention to (and have it already highlighted for your reader).
You know how to do that? For example, see here. ?

 

[BigBeachBanana]

Hmmm, see Post #2. ?

See post #4 for TLDR.

 

[The Highlander]

TLDR? ?
Post #2 was only eight lines long! ?‍
And right in the middle of it (in BIG print) it said:-

\(\displaystyle x^2+x^2\ne(x+x)^2\qquad(x+x)^2=4x^2~ \textbf{ not }~2x^2\)!!!​