Maximum Area of Rectangle in a Right Triangle - Problem with Solution

Maximize the area of a rectangle inscribed in right triangle using the first derivative. The problem and its solution are presented

Problem:
BDEF is a rectangle inscribed in the right triangle ABC whose side lengths are 40 and 30. Find the dimemsions of the rectangle BDEF so that its area is maximum.

Rectangle inscribed in right triangle problem

Solution to Problem:

  • let the length BF of the rectangle be y and the width BD be x. The area of the right triangle is given by (1/2)*40*30 = 600. But the area of the right triangle may also be calculated as the sum of the areas of triangle BEC and BEA. Hence

    600 = (1/2)*40*y + (1/2)*30*x

  • Let A be the area of the rectangle. Hence

    A = y*x

  • We now use the first equation to express y in terms of x as follows

    y = (600 - 15x) / 20

  • Substitute in A to obtain

    A(x) = x(600 - 15x) / 20

  • The graph of A(x) as a function of x is shown below. A(x) has a maximum value for x = 20. This will be shown analytically as well

    Rectangle inscribed in right triangle solution



  • An expansion of A(x) shows that A(x) is a quadratic function with negative leading coefficient and therefore has a maximum value.

    A(x) = -(3/4)x2 + 30x

  • We now calculate the first derivative of A.

    A'(x) = -(3/2)x + 30

  • Set A'(x) = 0 and solve for x.

    x = 20

  • It is easy to check that A'(x) is positive for x < 20 and negative for x > 20 and therefore A(x) has a maximum at x = 20. The maximum area is given by A(20)

    A(20) = -(3/4)202 + 30*20 = 300

  • We now find y as follows

    A = 300 = x * y , y = 300/20 = 15

  • The dimensions of the rectangle that make its area maximum are x= 20 and y = 15.

More references on calculus problems



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