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### Finding the Maximum Area of a Rectangle

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Date: 1/28/96 at 18:35:47
From: Anonymous

The sides of a triangle are 10cm, 10cm and 12cm.  A rectangle is
inscribed in the triangle, with one side on the 12cm side of the
triangle.  The other two vertices of the rectangle are on the 10cm
sides.  Find the maximum area of the rectangle.

I know the question involves coming up with an equation for the
area
of the rectangle, then finding the derivative and finding the
maximum value, etc... but what I don't know is how to set up the
question.

Help!
```

```
Date: 5/29/96 at 13:9:0
From: Doctor Anthony

Draw the perpendicular from the vertex where the two 10 cm sides
meet to the centre of the 12 cm side. The two right-angled
triangles so formed will be of dimensions, 6, 8, 10 cm - so the
height of the original isosceles triangle is 8 cm.  Let the
distance from the centre of the base to the vertical edge of the
inscribed rectangle be x, and the height of the rectangle be y.
The area of rectangle is A = 2xy.

Now by similar triangles we have y/(6-x) = (8-y)/x
This simplifies to xy = 48 - 8x - 6y + xy

4x + 3y + 24   or 3y = 24 - 4x

y = 8-(4/3)x

Then A = 2x{8-(4/3)x}
A = 16x - (8/3)x^2   Now differentiate

dA/dx = 16 - (16/3)x = 0 for max. or min.

Solve this to get x = 3.  Then y = 8 - (4/3).3
= 8 - 4
y = 4

And maximum area = 2xy = 2 x 3 x 4 = 24 cm^2

-Doctor Anthony,  The Math Forum

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Associated Topics:
High School Calculus

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