SEO - Optimization problem, finding the maximum area of a rectangle inside of a right triangle?

Optimization problem, finding the maximum area of a rectangle inside of a right triangle?

Another optimization problem that my professor didn't go over. Any help would be much appreciated guys, I don't even know where to start. The problem is:

Find the rectangle of maximum area that can be inscribed in a right triangle with legs of length 3 and 4 if the sides of the rectangle are parallel to the legs of the triangle.

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I won't give you the answer, but take a sheet of paper, draw a right triangle with side lengths 3 (y-axis) and 4 (x-axis). On the y-axis, mark points 0.5, 1, 1.5, 2, 2.5. From each of these points, draw a line parallel to the x-axis until they touch the hypotenuse. From this intersection point, draw a line parallel to the y-axis until it intersects the x-axis. You now have a rectangle. Measure the area of these rectangles. That should get you started and you should be able to generalize it from there.
I have a triangle a, b, c

a

.

.

.

c.................................b

The right angle is at c & ac = 3 and bc = 4

if I draw a line parallel to cb that intersects ac I get something that looks like this with p & w being the intersection points with the triangle:

a

.

p..................w

.

c.................................b

dropping a line from w to cb will create the rectangle you want to maximize

let's say the length of ap is x, then pc is 3 - x

Now all we need to know is the length of pw

well I know that the ratio of ap to ac is equal the ratio of pw to cb:

ap/ac = pw/cb --------fill in the known values to get:

x/3 = pw/4 ------------- solve for pw interms of x

4/3 x = pw

So, the area of the rectangle that you want to maximize is

(3-x) * 4/3 x = Area

I assume that you know how to maximize the given equation
Area = A = -4/3x^2+4x

dA/dx = -8/3x + 4 = 0 --> x = 1.5

y = -4/3(1.5) +4 = 2

Dimensions are 1.5 by 2 so area = 3units^2
Let ΔABC be a right triangle with ∠ A = 90°.

Let ADEF be the rectangle with D on AB, E on BC and F on AC.

Let BD = x and DE = y.

=> AD = 3 - x (taking AB = 3)

Area of rectangle

= DE * AD

= y (3 - x)

By similarity of ΔCFE, ΔEDBand CAB,

(3 - x) / (4 - y) = x / y = 3 /4

=> 3 - x = (3/4) (4 - y)

=> area of rectangle

= (3/4) y * (4 - y)

= (3/4) (4y - y^2)

= (3/4) [4 - (y -2)^2]

This is minimum when y - 2 = 0 and the minimum

area of the rectangle is

(3/4) x 4

= 3.

[Note: This is half the area of the right tringle. In fact, it can be proved that the rectangle inscribed in any triangle will have area which is half the area of the triangle.]