SOLUTION: How to find the Maximum Area of the Rectangle inscribed in a Right Triangle of sides a, b and Hypotenuse length c. Please I want a Algebra Proof, not one using Calculus or Derivati
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Question 985127: How to find the Maximum Area of the Rectangle inscribed in a Right Triangle of sides a, b and Hypotenuse length c. Please I want a Algebra Proof, not one using Calculus or Derivatives.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
.
Refer to the figure.
The area of right triangle ABC is 
. But the area of the right triangle is also the sum of the areas of the two triangles formed by the diagonal of the inscribed rectangle, namely triangle CEB and CEA. The side of the rectangle that measures 
is the altitude of triangle CEB and the side of the rectangle that measures 
is the altitude of triangle CEA. So the areas of these two triangles are given by 
and 
. Hence, the following relationship holds for any arbitrary right triangle with an inscribed rectangle:
Solving for y:
(you can verify the algebra for yourself)
Then, since the area of a rectangle is given by the length times the width, the area of the rectangle as a function of is:
\ =\ ax\ -\ \frac{ax^2}{b})
or, in standard form:
\ =\ -\ \frac{ax^2}{b}\ +\ ax)
Note that the graph of this function is a parabola that opens downward because of the negative lead coefficient.
The -coordinate of the vertex of a parabola of the form \ =\ \alpha{x^2}\ +\ \beta{x}\ +\ \gamma)
is given by
For your area function, 
and 
, hence the -coordinate of the vertex, and therefore the point where the maximum value of the area function will be found is at:
The maximum rectangle is formed when the sides of the rectangle are exactly one-half of the legs of the right triangle.
John
My calculator said it, I believe it, that settles it
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