Finding the Maximum Area of a RectangleDate: 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 area 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 question. Help! 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 |
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