Maximum and Minimum (1 Viewer)

Roserox · Nov 30, 2015

A piece of wire of length 5 metres is bent to form the hypotenuse and a side of a right angled triangle ABC. Let the length of the side AB be x metres. What is the maximum possible area of the triangle?

Thank you for any help in advance!

InteGrand · Nov 30, 2015

Roserox said:
A piece of wire of length 5 metres is bent to form the hypotenuse and a side of a right angled triangle ABC. Let the length of the side AB be x metres. What is the maximum possible area of the triangle?

Thank you for any help in advance!

$Here's a start. Let the other (non-hypotenuse) side be length $y$ metres, then we have the hypotenuse being $\sqrt{x^2 + y^2}$ (Pythagoras' Theorem). So we must have $x+y+\sqrt{x^2 + y^2} = 5$ (as the total length of the wire is 5 metres). This is our constraint equation. The function to optimise (the objective function) is $A = \frac{1}{2}xy$. So we have set up our constraint equation and objective function, so what remains is to convert the problem into a one-variable optimisation problem using the constraint equation as is usual for these questions.$

Paradoxica · Nov 30, 2015

Roserox said:
A piece of wire of length 5 metres is bent to form the hypotenuse and a side of a right angled triangle ABC. Let the length of the side AB be x metres. What is the maximum possible area of the triangle?

Thank you for any help in advance!

InteGrand said:
$Here's a start. Let the other (non-hypotenuse) side be length $y$ metres, then we have the hypotenuse being $\sqrt{x^2 + y^2}$ (Pythagoras' Theorem). So we must have $x+y+\sqrt{x^2 + y^2} = 5$ (as the total length of the wire is 5 metres). This is our constraint equation. The function to optimise (the objective function) is $A = \frac{1}{2}xy$. So we have set up our constraint equation and objective function, so what remains is to convert the problem into a one-variable optimisation problem using the constraint equation as is usual for these questions.$

$Using the AM-GM we have:$

$x + y \geq 2\sqrt{xy}$

$x^2 + y^2 \geq 2xy$

$\noindent Since $x$ and $y$ are positive, we are allowed to apply AM-GM to this scenario. Take the square root of the second inequality, which is valid, as both sides are non-negative.$

$\sqrt{x^2 + y^2} \geq \sqrt{2xy}$

$Add both inequalities together to yield:$

$x + y + \sqrt{x^2+y^2} \geq 2\sqrt{xy} + \sqrt{2xy}$

$Factorise the RHS and divide away the constant term.$

$\frac{x + y + \sqrt{x^2+y^2}}{2+\sqrt{2}} \geq \sqrt{xy}$

$From Integrand's post we have 5 on the LHS and A on the RHS$

$\frac{5}{2+\sqrt{2}} \geq \sqrt{2A}$

$The rest can be cleaned up by squaring everything and a bit of re-arrangement to yield:$

$A \leq \frac{25}{4}(3-2\sqrt{2})$

$Hence, the maximum possible area of the triangle is approximately $ 1.07223 $ square units.$

leehuan · Nov 30, 2015

You did not just quote the AM-GM inequality in 2U

Paradoxica · Nov 30, 2015

leehuan said:
You did not just quote the AM-GM inequality in 2U

In theory, I could have proven it using the trivial inequality but cbbs.

leehuan · Nov 30, 2015

It's 2U. You might even want to just briefly explain why (x-y)^2>=0 before going through all of that. Not even a 3U student is expected to be able to do that proof.

Paradoxica · Dec 1, 2015

leehuan said:
It's 2U. You might even want to just briefly explain why (x-y)^2>=0 before going through all of that. Not even a 3U student is expected to be able to do that proof.

Well that one is easy, just use calculus treating y as a constant.

InteGrand · Dec 1, 2015

Paradoxica said:
Well that one is easy, just use calculus treating y as a constant.

Or more simply, just state that the square of any real number is non-negative; this would be sufficient in an exam.

Maximum and Minimum (1 Viewer)

Roserox

New Member

InteGrand

Well-Known Member

Paradoxica

-insert title here-

leehuan

Well-Known Member

Paradoxica

-insert title here-

leehuan

Well-Known Member

Paradoxica

-insert title here-

InteGrand

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)