A rectangle is inscribed in an equilateral triangle so that one side of the rectangle lies on the base of the triangle. How do I find the maximum area of the rectangle when the triangle has side length of 10?

1 Answer
Dec 24, 2015

#A = (25sqrt(3))/2#

Explanation:

First, let's look at a picture.

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Some initial observations:

  • The area #A# of the rectangle is #A=bh#.

  • By symmetry, the base of the triangle is of length #b+2t#, and thus, as it is of length #10#, we have #b+2t = 10 => t = 5-b/2#

  • If we decide #b# that also determines #h#, and thus we can write #h# as a function of #b#.

To write #h# as a function of #b#, we can look at the right triangle with legs #t# and #h#. As it shares an angle with an equilateral triangle, we know it is a #30^@#-#60^@#-#90^@# triangle, and thus #t/h = 1/sqrt(3)#.
Solving for #h# gives us #h = sqrt(3)t = sqrt(3)(5-b/2)# by our initial observation.

Then, we can rewrite our formula for the area as

#A = b*sqrt3(5-b/2) = sqrt(3)(-1/2b^2 + 5b)#

If we look at the graph for #A# we will see it is a downward facing parabola, and thus will have a maximum at its vertex. Then, we can complete the square to find

#A = sqrt(3)(-1/2b^2+5b)#
# = -sqrt(3)/2(b^2-10b)#
# = -sqrt(3)/2((b-5)^2-25)#

And so the vertex, and thus the maximum area, is at #b = 5#.

Finally, we calculate the area from this to get

#A = -sqrt(3)/2((5-5)^2-25) = (25sqrt(3))/2#