Asked by Anonymous
Find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 4 cm and 6 cm if two sides of the rectangle lie along the legs.
Answers
Answered by
mathhelper
let the base of the rectangle be x
let the height of the rectangle be y
If you make a sketch you can see three similar triangles,
by ratios:
(6-x)/y = 6/4
6y = 24 - 4x
y = (12 - 2x)/3
area = xy
= x((12-2x)/3 = 4x - (2/3)x^2
d(area)/dx = 4 - (4/3)x
= 0 for a max of area
(4/3)x = 4
4x = 12
x = 3, then y = (12-6)/3 = 2
So the largest rectangle has an area of 6 units^2
let the height of the rectangle be y
If you make a sketch you can see three similar triangles,
by ratios:
(6-x)/y = 6/4
6y = 24 - 4x
y = (12 - 2x)/3
area = xy
= x((12-2x)/3 = 4x - (2/3)x^2
d(area)/dx = 4 - (4/3)x
= 0 for a max of area
(4/3)x = 4
4x = 12
x = 3, then y = (12-6)/3 = 2
So the largest rectangle has an area of 6 units^2
Answered by
Anonymous
6 on bottom
4 up
w horizontal
h up from 6 side
then
w/6 = (4-h)/4
or
h/4 = (6-w)/6 (same thing by similar triangles)
Area = (1/2) w h = A
A =(1/2) w (4 - 2 w/3) = 2 w - w^2/3
dA/dw = 0 at max/min = 2 - 2 w/3
w = 3
then h = 4 - 2(3)/3 = 2
4 up
w horizontal
h up from 6 side
then
w/6 = (4-h)/4
or
h/4 = (6-w)/6 (same thing by similar triangles)
Area = (1/2) w h = A
A =(1/2) w (4 - 2 w/3) = 2 w - w^2/3
dA/dw = 0 at max/min = 2 - 2 w/3
w = 3
then h = 4 - 2(3)/3 = 2
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