Given the triangle OAB, O(0,0), A(10,0),B(6,12),find the dimensions of the rectangle inscrjibed in triangle, with one side on OA ?The rectangle has...

The vertex found on OB is M and we notice that the segment
OM has the same slope as the side of triangle, OB.

We'll
calculate the slope of OB:

mOB = (yB - yO)/(xB -
xO)

mOB = (12-0)/(6-0)

mOB =
12/6

mOB = 2

So, the slope of
OM has the same value, mOM =2. The coordinates of
M(x,y).

mOM = (y -
0)/(x-0)

But mOM = mOB = 2

2 =
y/x

y = 2x (1)

The width of
the rectangle is W = 2x.

The vertex found on AB is N and we
notice that the segment AN has the same slope as the side of triangle,
AB.

We'll calculate the slope of
AB:

mAB = (yB - yA)/(xB -
xA)

mAB = (12-0)/(6-10)

mAB =
12/-4

mAB = -3

So, the slope
of AN has the same value, mAN =-3. The coordinates of
N(x,y).

mAN = (y -
0)/(x-10)

But mAN = mAB =
-3

-3 = y/(x-10)

y =
-3(x-10)

We'll substitute x by x +
L:

y = -3(x + L - 10)
(2)

We'll equate the expressions (1) and (2) to determine
L:

2x = -3(x + L - 10)

We'll
remove the brackets:

2x = -3x - 3L +
30

3L = -5x + 30

L = -5x/3 +
10

The length of the rectangle is L = -5x/3 +
10

We'll calculate the area of the inscribed
rectangle:

A = L*W

A = (-5x/3
+ 10)*2x

We'll remove the
brackets:

A = -10x^2/3 +
20x

We notice that the function of the area is a quadratic
whose leading coefficient is negative, so the parabola will have a maximum point for the
critical point of the function.

We'll determine the
critical point, differentiating the function:

A'(x) =
-20x/3 + 20

We'll put A'(x) =
0

 -20x/3 + 20 = 0

We'll
divide by 20:

-x/3 + 1 =
0

-x/3 = -1

x =
3

The critical point of the function is x = 3, for the
function has a maximum point:

A(3) = -10*3*3/3 +
20*3

A(3) = -30 + 60

A(3) =
30

The maximum value of the area of the
rectangle inscribed in the given triangle is A = 30 square
units.