# Thread: Maximizing/Minimizing the area of a rectangle

1. ## Maximizing/Minimizing the area of a rectangle

HOw can you maximize/minimize the area of a rectangle inscribed in a triangle?
See the figure below w/ given measurements of the triangle.
10 , 8 and 6

2. Originally Posted by ^_^Engineer_Adam^_^
HOw can you maximize/minimize the area of a rectangle inscribed in a triangle?
See the figure below w/ given measurements of the triangle.
10 , 8 and 6
Choose a single parameter which describes the size of the rectangle (see the attachment), and calculate the area of the rectangle in terms of that single parameter.

So here:

A=(3/4) x(8-x)

which you are to find the maximum of (for x in [0,8]).

RonL

3. Originally Posted by CaptainBlack
So here:

A=(3/4) x(8-x)

which you are to find the maximum of (for x in [0,8]).

RonL
We do this by differentiating A with respect to x, setting the derivative to zero, and solving for x. This is the x which gives the maximum (or other stationary point type, but in this case it is a maximum) area.

A=(3/4) (8x-x^2)

dA/dx=(3/4)(8-2x)

so dA/dx=0, gives 2x=8, or x=4.

Plugging x=4 back into the formula for the area gives the maximum area of a rectangle inscribed in the given triangle is:

A=(3/4) 4(8-4)=12.

and the actual dimensions of the rectangle can be found from the relations between u, v x and the dimensions of the triangle shown in the diagram in my previous post

RonL

4. Originally Posted by ^_^Engineer_Adam^_^
HOw can you maximize/minimize the area of a rectangle inscribed in a triangle?
See the figure below w/ given measurements of the triangle.
10 , 8 and 6
Here is one way.
We will use your figure as posted.

Let x = width of the inscribed rectangle
And y = length.

On the figure, on the horizontal leg of the 6-8-10 right triangle, or on the 6-unit long leg. Let's call that horizontal leg, line segment AB, where
point A is the corner of the big triangle at the right angle
point C is the corner of the big triangle, at the other end of the AB.

Call the 3rd corner of the big triangle, the one at the top, as point C.
Call the corner of the inscribed rectangle along CA as point D.
Call the corner of the inscribed rectangle along AB as point E.
Call the lower corner of the inscribed rectangle along BC as point F.

Right triangles DAE and CAB are similar, so, are proportional.
DE/CB = AE/AB
y/10 = AE/6
AE = 6y/10 = 3y/5 ------------**

Right triangles EFB and CAB are similar, so, are proportional.
EB/CB = EF/CA
EB/10 = x/8
EB = 10x/8 = 5x/4 ----------**

Now,
AE +EB = AB
3y/5 +5x/4 = 6
Clear the fractions, multiply both sides by 5*4,
12y +25x = 120
12y = 120 -25x
y = (120 -25x)/12
y = 10 -25x/12 ---------------(i)

Area of inscribed rectangle, A = x*y
A = x(10 -25x/12)
A = 10x -(25x^2)/12
Differentiate both sides with respect to x,
dA/dx = 10 -(1/12)(50x)
dA/dx = 10 -25x/6
Set dA/dx to zero,
0 = 10 -25x/6
x = 10 / (25/6)
x = 60/25 = 12/5 = 2.4 units long -------------for max A.
So,
y = 10 -25x/12 = 10 -25*2.4/12 = 10 -5 = 5 units long ----for max A.

Therefore, to maximize the area of the inscribed rectangle, positioned as shown on the figure, make the width = 2.4 units long, and the length = 5 units long. -----------------answer.

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### how to maximise minimize inscribed figure

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