# 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 maximum area.

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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.**