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Finding the Maximum Area of a Rectangle

Date: 1/28/96 at 18:35:47
From: Anonymous
Subject: Grade 12 Calculus

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 
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 


Date: 5/29/96 at 13:9:0
From: Doctor Anthony
Subject: Re: Grade 12 Calculus

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

Associated Topics:
High School Calculus

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