Proof by area of maximisation (1 Viewer)

[z600]

A rectangle is inscribed in a isoceles triangle. Prove that when the area of the rectangle is maximised, its area is half of the isoceles triangle.

thanks

 

[getridofns]

we first label any generic isosceles triangle on the coordinate plane, letting it have vertices . Then we inscribe a rectangle into the isosceles triangle and let it have coordinates (we obtain the last two points by finding the equation of the line connecting and and then by plugging in ). Hence we know that the area of the triangle is and the area of the rectangle is . We wish to prove that:



simplifying this we obtain:



therefore we know the area of the rectangle is maximized when the equality holds and

hence we know which means that

and so the area of the rectangle is which is half that of the triangle
p.s what book this qn in??

 

[z600]

we first label any generic isosceles triangle on the coordinate plane, letting it have vertices . Then we inscribe a rectangle into the isosceles triangle and let it have coordinates (we obtain the last two points by finding the equation of the line connecting and and then by plugging in ). Hence we know that the area of the triangle is and the area of the rectangle is . We wish to prove that:



simplifying this we obtain:



therefore we know the area of the rectangle is maximized when the equality holds and

hence we know which means that

and so the area of the rectangle is which is half that of the triangle
p.s what book this qn in??

i solved it by similar triangles. its in cambridge

 

[m.incognito]

Okay tell me that 2010'er did NOT just answer the question.
WHAT A FREAK to let alone know what it was on about.

FRICKIN...

I mean I do extension, and I'm just ???

 

[z600]

Okay tell me that 2010'er did NOT just answer the question.
WHAT A FREAK to let alone know what it was on about.

FRICKIN...

I mean I do extension, and I'm just ???

LOL jealous?

 

[m.incognito]

Yes I really am
I know. I fail. =[

 

[getridofns]

How did you do through similar triangles.. and i see my latex didnt show up

p.s im in year 12.. i needed to make an account quick so i clicked 2010

 

[getridofns]

we first label any generic isosceles triangle on the coordinate plane, letting it have vertices

. Then we inscribe a rectangle into the isosceles triangle and let it have coordinates

(we obtain the last two points by finding the equation of the line connecting

and

and then by plugging in

). Hence we know that the area of the triangle is

and the area of the rectangle is

. We wish to prove that:

simplifying this we obtain:

therefore we know the area of the rectangle is maximized when the equality holds and

hence we know

which means that

and so the area of the rectangle is

which is half that of the triangle


--------------------------------------

this is a bloody good qn... i think ill note it down

thanks

 

[getridofns]

Proof for any triangle...

We have two cases for a rectangle being inscribed within a rectangle.

Case 1 (see Figure 1)

The triangle is right angled, and the inscribed rectangle has a vertex at this right angle.

Case 2 (see Figure 2)

One side of the rectangle lies upon one side of the triangle (think of this as sharing four of the rectangle's vertices between three of the triangles sides). The altitude which is perpendicular to this side always passes through the rectangle. If this were not the case then the four rectangle's vertices would have to be shared between two of the triangle's sides. This is only possible when these two sides are parallel - an obvious contradiction.

Lets begin with Case 1

By similar triangles:

If we were to sketch a graph of

against

a concave down parabola would be obtained which intersects the

-axis at

and

. The maximum value of

is found midway between these roots so corresponds to

Hence

Now lets turn to Case 2:

By similar triangles:

This is the same expression for

as in Case 1, and by applying the same logic, the maximum rectangle which can be inscribed is half the area of the triangle. QED.

 

[getridofns]

is this an extension or normal question??