July 2002:
Home | Problem Archive | Past Winners | Number Games | Contact Me
07/22/02) An equilateral trapezoid is inscribed in a parabola of equation y = k-x2 where one of its bases lie on the x-axis from one x-intercept to the other. Find the maximum area it could have in terms of k, where k is a positive constant.
Let us start by finding the bases and height of the trapezoid in terms of k. As we can calculate, the x-intercepts of the parabola would be at (sqrt(k),0) and (-sqrt(k),0). The bottom base would be simply 2sqrt(k). The top base would be 2x where x is the x-coordinate at which the area is maximum. The height is simply the y-value or k-x2. Now we combine these into a single equation for the area which is generally (b1+b2)*h/2. The area in terms of k is then (k-x2)(x+sqrt(k)). For the area to be maximum, the derivative of area with respect to x has to be zero. Therefore, we set -3x2-2x*sqrt(k)+k = 0. Using the quadratic equation this would hold when x is sqrt(k)/3 making y = 8k/9. Hence, the maximum area is 32k*sqrt(k)/27 or 32k3/2/27. Cool, isn't it?
07/29/02) A parabola goes through the points (7,0) and (-3,164) and at x = 4, the slope of its tangent line is 17. Find the y-coordinate of its y-intercept and the x-coordinate of its vertex.
The clue is parabola. That tells you right away to use the quadratic formula, which is y = ax2+bx+c. Two points are given, so that's two equations when substituting y and x. We need one more equation, because we have three unknowns a,b and c. But, we have the derivative of the parabola at x = 4. That means 2ax+b = 17 when x is 4. That adds on to our equations. Using any method for solving the three equations, we come up with the formula for the parabola being y = 12x2-79x-35. Therefore, the y-coordinate of its y-intercept is -35 and the x-coordinate of its vertex is 79/24. Sweet!
______________________________________
07/15/02) Does the sum of (n!*en)/(nn*2n) from n = 1 to infinity converge or diverge? Prove it by a test.
The best way is to apply the Ratio Test and find out if the ratio is less or greater than 1. If equal 1, the test is inconclusive. Doing so, we divide the expression using n+1 instead of n by the expression with n and take the limit as n goes to infinity. After taking the limits and simplifying the algebra, it is amazing that the ratio is 1/2 hence making the series converge.
______________________________________
07/08/02) A parabola has a vertex at (0,4) and touches the x-axis. Its first-quadrant region's area is equal to the volume produced by rotating that area about the y-axis. Find its exact formula and the value of the volume.
Let us start by making a general equation for the situation. The parabola can only have an equation of the form y = kx2+4. The first-quadrant x-intercept would then be the square root of -4/k. Now we set up the equations. The integral of kx2+4 from x = 0 to x = sqrt(-4/k) would be equal to the integral of pi*x2 from y = 0 to y = 4. In this case, x is solved in terms of y and then squared. If we go ahead and solve the equations, it would yield to k = -9pi2/4, which is approximately -22.21 and the volume - or area - would be 32/9pi or about 1.13. Wooh, a lot of work, wasn't it?
______________________________________
07/01/02) In a party, you can choose between two glasses, in which you can drink. One has a parabolic shape, the other a linear. Yet, both have the same height of 15 cm. If the parabolic glass is 8 cm in diameter on the top, 6 cm in diameter at half of its height and 3 cm in diameter on the bottom, whereas the linear one is 9 cm in diameter on top and 4 cm on the bottom. Comparing the volumes of each, which one should fill up completely and drink in order not to exceed a limit of 40 mL?
Woo, a lot of information. Let us first investigate for a way to set up coordinates and make up equations for our glasses' shape. Doing this, we need to use points (2,0) and (9/2,15) for the linear glass and points (3/2,0), (3,15/2) and (4,15) for the parabolic one. These are obtained using the radii and heights that are given in the problem. Now, let us find equations. The general formulas for a line and a parabola eventually lead us to the linear equation of y = 6x-12 and the quadratic equation of y = x2+x/2-3. Not done yet! We have to find the volumes. One way is to use disks of radius x. This would get us a volume of about 36.69 mL for the parabolic glass and 50.85 mL for the linear one. Therefore, after all these calculations, we come up with the fact that the parabolic should be use. Don't drink and drive, no matter which one is chosen!!
| Previous Month | Archive | Next Month |
|---|