Answer:
x = 2 cm
y = 2 cm
A(max) = 4 cm²
Step-by-step explanation: See Annex
The right isosceles triangle has two 45° angles and the right angle.
tan 45° = 1 = x / 4 - y or x = 4 - y y = 4 - x
A(r) = x* y
Area of the rectangle as a function of x
A(x) = x * ( 4 - x ) A(x) = 4*x - x²
Tacking derivatives on both sides of the equation:
A´(x) = 4 - 2*x A´(x) = 0 4 - 2*x = 0
2*x = 4
x = 2 cm
And y = 4 - 2 = 2 cm
The rectangle of maximum area result to be a square of side 2 cm
A(max) = 2*2 = 4 cm²
To find out if A(x) has a maximum in the point x = 2
We get the second derivative
A´´(x) = -2 A´´(x) < 0 then A(x) has a maximum at x = 2
The maximum area of the rectangle inscribe in the right isosceles triangle
is [tex]\rm \bold{4\;cm^2}[/tex]
Given that a rectangle is to be inscribed in an isosceles triangle
As per the attached figure we can write the following
Let ΔABC represents the isosceles triangle from the given data in the question we can write the following observations
Length of AB = 4
Length of BC = 4
∠ACB = 45°
∠BAC = 45°
Length of AC = [tex]\rm 4\sqrt2[/tex]
We have to consider a rectangle within the isosceles triangle such that
area of rectangle is maximum
The area of rectangle is given by equation (1)
[tex]\rm Area \; of \; rectangle = Length \times Breadth ........(1) \\[/tex]
Let the area of rectangle inscribed by the isosceles triangle is " A"
Let the lenght of the rectangle be "x "
Let the width of the rectangle be " y "
So according to equation (1) the area of rectangle inscribed into the triangle is given by
[tex]\rm A = x \times y.......(2)[/tex]
In ΔAED we can write
[tex]\int\limits^a_b {x} \, dx \rm tan 45= \dfrac{Perpendicular}{Base } = \dfrac{4-y}{x} \\1 = \dfrac{4-y}{x} \\ x = 4-y .......(3) \\[/tex]
From equation (2) and (3) we get
[tex]\rm A = (4-y)(y) \\A = 4y -y^2......(4)[/tex]
On differentiating the given function with respect to y we get
[tex]\rm A' = 4- 2y \\[/tex]
On putting A' =0 we get
[tex]\rm 4-2y =0 \\2y = 4\\y = 2[/tex]
A'' is negative hence it is the maximum value of y
So we can conclude that the maximum value of y is 2
So from equation (4) the maximum area is [tex]\rm 2\times 2 = 4\; cm^2[/tex]
The maximum area of the rectangle inscribe in the right isosceles triangle
is [tex]\rm \bold{4\;cm^2}[/tex]
For more information please refer to the link given below
https://brainly.com/question/541676
