January 2002:

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01/28/02) If f(x+1)=x2-5x, what is the local extrema of f(x)?

Solution:

The question is referring to the following series:
More than a calculus problem, this is an algebra (advanced topics) problem. F(x+1) should not be confused with f(x), because any algebraic work has to be operated on x+1. To get x2-5x it is necessary first to square the quantity x+1. So we are left with x2+2x+1. Now to get -5x, 7x has to be subtracted. But REMEMBER that we only work with x+1. So we add -7(x+1). Now 1-7 is -6, so we add +6 to cancel that. Therefore, we get:
f(x+1)=(x+1)2-7(x+1)+6. More generally,
f(x)=x2-7x+6.
Now in the calculus part, we set its derivative equal to zero and we get 2x-7=0. So the answer is x=7/2. Sweet mathematics, right?

Topic:

Advanced Topics

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01/21/02) The opposite of the square of a number is chosen. It is then added to its half, again added to the result's half and so on. What is the input that makes the result of this process maximized?

Solution:

The question is referring to the following series:
-x2 + (-x2/2) + (-x2/4) + ...
According to geometric series formulas, if r, ratio of multiplication is less than 1, the sum is t1/(1-r). In this particular case, it is
-x2/(1-1/2), which is -2x2.
Now, to maximize this result, we have to set its derivative equal to zero. So -4x=0. Therefore, the required input is 0. Incredible, isn't it?

Topic:

Series and Derivatives

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01/14/02) What is the largest number that exceeds its square? Its cube? Its nth power?

Solution:

By setting up an equation for this problem, we obtain x-x2, and taking its derivative, we get 1-2x and solving it for x when the derivative is zero, the answer is 1/2. The cubic one is also the same process, so x is square root of 1/3 or root3/3. The nth power can be solved in the same way too. x-xn is the function and 1-nxn-1 is the derivative. So the largest number that exceeds its nth power, is (n-1)th root of 1/n. Very interesting!

Topic:

Optimization

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01/07/02) A rectangular cube is to be made in such a way that one side is x, the other is 1m less than x and the last side is 3m bigger than x. The cube has to have the lowest volume possible. What should the sides be to fulfill this requirement?

Solution:

Ironically, there is no solution for this question. We first have to put the problem in mathematical terms. So volume is x(x-1)(x+3), and V'(x), which is the rate of change of the volume, is 3x2+4x-2. By setting this equal to zero, we get two values for x, which are approximately 0.535 and -1.87. However, x cannot be less than 1, because that would make the volume negative, something that does NOT make sense. Therefore, there is no solution to this problem.

Topic:

Optimization

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