Max and Min (1 Viewer)

[Lukybear]

Ahhh This is driving me crazy.

A rectangle PQRS is placed inside the scalene triange ABC as shown. If the area of Triangle ABC is constant, prove that the maximum area of the rectangle is one-half (1/2) the area of triangle ABC.

 

[Lukybear]

bump?

 

[Lukybear]

Any1?

 

[bob fossil]

get the 2 equations. subsititute to get rid of varibales. use first derivative ......

 

[Lukybear]

O jeez. Get me a hint?

 

[bob fossil]

make an equation for the triangle using x and y.
make an equation for the retangle using x and y.
make y the subject of one of them.
sub y into the other.

thats how you do things normaly

 

[Lukybear]

Cmon... someones gota know?

 

[Lukybear]

Please, I am desperate!

 

[mecramarathon]

use similar triangles - is this Q from 2U fitzpatrick?

h = height of triangle b = length of triangle w = width of rectangle l = length of rectangle

anyway (h - w) / h = L / b (corr. sides of similar triangle)

therefore w = h - hL/b

Area of rectangle = L x W
= L ( h - hL/b)
= hL - hL*2/b
Differentiate DA/DL = h - 2hL/b
h - 2hL/b = 0
2hL/b = h
L = b/2 (cancel the h)

Therefore Area of rectangle = h x b/2 - h/b x b*2/4
= bh/2 - bh/4
= 1/2 x 1/2 bh
= half area of triangle

apparently a lot of schools use this q in their trial papers

 

[addikaye03]

make an equation for the triangle using x and y.
make an equation for the retangle using x and y.
make y the subject of one of them.
sub y into the other.

thats how you do things normaly

ROFL! Helpful man.

Sorry i didn't see Q earlier OP. Above is correct, welldone mecramarathon

 

[Lukybear]

Darn... Cant believe i missed it. Thx so much, thats been teasing me all day now.