The Tennis Blog

Have a few annoying calculus questions, help would be awesome.?

1.) Find the equation of a given parabola that has a maximum value of 5 if the parabola is in the form of ax²+3x-4

1) First you need to find the derivative of the expression. Using f(x) to represent the function of x, you have:
f(x) = ax^2 + 3x – 4
df/dx = f’(x) = 2ax + 3

An extremum, either a maximum or a minimum, can be found be setting the derivative equal to zero (i.e. f’(x) = 0 ). With a parabola you will have only one extremum, and that extremum will be either a maximum or a minimum. Using the derivative we have:
2ax + 3 = 0.

For simplicity here I am assuming that when it states that it has a “maximum value of 5″ that the question means that the maximum occurs when x = 5. (The correct interpretation is that the maximum occurs at f(x) = 5. Without knowing the value of a yet this is a bit messy, but can still be done. When solving this question remember to solve for x when f(x) = 5 and then using that value for x instead of 5 as I have done.)

Since we are assuming x = 5:
2*5a + 3 = 0
10a + 3 = 0
a = -3/10 (Solve for a)

Which would make the equation:
(-3/10)x^2 + 3x – 4

–>Remember this is not your answer. This is the answer I am using for simplicity of arithmetic (typing/reading typed lengthy fractions is painful). You should use f(x) = 5 as the maximum rather than x = 5 (the difference between “maximum value of 5″ and “maximum value occurs at 5″), as indicated previously.

2) From what I recall when you are asked to “complete the square” you are being asked to make a quadratic equation into a square. So let’s start with an example with a square:
(x-1)^2

This can be shown to be:
(x-1)^2 = x^2 – 2x + 1

If you were given:
x^2 – 2x + 2 = 0

And you were asked to “complete the square”, you would set a quadratic that is a perfect square equal to something. Here:
x^2 – 2x + 2 = x^2 – 2x + 1 + 1 = 0

Then:
x^2 – 2x + 1 = -1
(x-1)^2 = -1
-> (x-2)^2 – 1 = 0

I’m not really sure how you would do this in your particular case since you do not have a specific equation (a, b, and c could already be such that the equation IS a perfect square, since they are not defined). On a side note, typically I’m not a fan of being asked to solve a question a particular way (even though I understand that it is necessary to practice a technique). I just ask a student I’m tutoring to “solve the question using whatever tools you know”. This unfortunately won’t help too much, but I will solve it so you can check your answer.

Recall how to find an extremum (for a parabola this is the vertex) from question (1). So:
y’(x) = 2ax + b

The vertex is:
2ax + b = 0

Solving for x:
x = -b/2a (so long as a != 0)

Now you substitute this in to find y:
y(-b/2a) = a(-b/2a)^2 +b(-b/2a) + c
= c – b^2/4a

So your vertex would be:
(x,y) = (-b/2a, c – b^2/4a)

3) Let’s let x = the number of tennis rackets, c = the cost of the tennis rackets, and y = total income. In general you will have:
y = cx

So if the cost is $100, you have c = 100:
y = 100x

Using this equation when you sell 600 tennis rackets, x = 600 and your income is:
y = 100*600 = 60 000

Or $60 000.

But here we have c as a function of x since as c changes, so does x:
x + 50 = c – 5
x = c – 55
(Note: You solve for x since you are looking for the price/cost, c, that will bring the largest monthly income – so you must eliminate x from your original equation.)

So:
y = cx = c(c – 55)
= c^2 – 55c

The largest monthly income would be the extremum (here a maximum). So now we need the derivative of y:
y’(c) = 2c – 55

The extremum occurs when the derivative is zero:
2c – 55 = 0

Solving for c:
2c = 55
c = 55/2 = 27.50

So if you charged $27.50 for each racket you would have the largest income. You can substitute this value of c into c(x) and y(c,x) to determine what your income will be (if necessary).

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