How do you find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 3 cm and 8 cm if two sides of the rectangle lie along the legs?

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Answer 1 · 2017-07-23T00:13:29

Its area is #6"cm"^2#

The largest possible rectangle must have a vertex that touches the hypotenuse of the triangle at one point.

Let us put the right angle of the triangle at the point #(0, 0)#, another vertex at #(8, 0)# and the final vertex at #(0, 3)#

We can represent points on the hypotenuse parametrically as:

#(8t, 3(1-t))#

where #t in [0, 1]#

Then the area of the rectangle with vertices:

#(0, 0)#, #(8t, 0)#, #(8t, 3(1-t))#, #(0, 3(1-t))# is:

#f(t) = 8t*3(1-t) = 24t(1-t) = 24(t-t^2) = 24(1/4-(t-1/2)^2)#

This takes its maximum value when #t=1/2# and #f(t) = 24*1/4 = 6#