May 2004:
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| May 2004: | |
| A rectangle is subscribed in a 30-60-90 triangle, so that one side of the rectangle resides on the triangle's hypotenuse, of length h. Calculate the maximum area of the rectangle in terms of h. Bonus: If the right triangle has an acute angle θ find the maximum area in terms of h and θ. |
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| Let the width of the rectangle be x. Using 30-60-90 triangular ratios of sides, the hypotenuse of the right triangle contains the length of the rectangle plus x*sqrt(3) and x/sqrt(3). Hence, the length of the rectangle in terms of x, its width, is h - 4x*sqrt(3)/3. This is good, because now we can express the area in terms of x, and take its derivative and set it equal to zero. With some algebra, this yields to x = sqrt(3)*h/8 and hence the maximum area is sqrt(3)*h2. To check for the extrema being a maximum, one can take the second derivate of the area and check for it to be negative, but it is intuitively clear. If the angle is θ instead of 30 degrees, the length of the rectangle becomes h - x*tan θ - x*cot θ. Doing the same procedure of setting the derivative of area equal to zero, we get x = h / (2tan θ + 2cot θ). Thus, the maximum area is equal to h2 / 4(tan θ + cot θ).
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