Remember when away from math for a period of time, we lose our skills to solve problems quickly. So it's always a good idea to review the basics in the math class you are currently taking or review the most recent material. When school begins again, you'll be ready to go and not fall behind. Math is an ongoing process, what is learned previously needs to be mastered in order to do well in the material that is to come.

Suppose you have 100 inches of wire to form the skeleton of a rectangular box with a square base. What dimensions of the box will maximize the volume?

The formula for volume for the box is area of the base times the height. Side the base is square, the area of the base is x^2, with x being the length of a side of the square. Let's make the height "y", therefore the volume is (x^2)y.

The distance around the base is 4x since each side is of length "x", therefore it is 8x around the top and bottom of the box. The sides are each length "y", so the four sides have length 4y.

So we know that 8x + 4y = 100.

Now solve the equation for either x or y. Solving for y, we get y = (100 - 8x)/4 = 25 - 2x.

Substitute that in for the volume equation to get volume = x^2(25 - 2x) = 25x^2 - 2x^3. To maximize the volume, take the derivative of the volume equation, which equals 50x - 6x^2.

Set this equal to zero and solve for x to get, 2x(25 - 3x) = 0. Therefore x = 0 or 25/3. 

The dimensions of the box are 25/3 by 25/3 by 25/3, so the box is a cube.

A ratio that is seen many times in geometry, architecture and art is the "golden ratio", which equals approximately 1.618.

We can find the golden ration by dividing a line into two parts such that the longer part, which we can label "a" divided by the smaller part, labeled "b" is also equal to the entire length of the line "a + b" divided by the longer part "a".

a/b + (a + b)/a

It is said that the most appealing rectangle to the eye is in the golden ratio.

One such rectangle is one with length 10 and width 6.18.

a = 10
b = 6.18

10/6.18 = 16.18/10 which is approximately 1.618

Suppose you have a function f(x) = 4x^3 + 3x^2 + 5x + 2.  What is the slope function and how do can you check that your slope function is correct?

The derivative of f(x), f'(x) = 12x^2 + 6x + 5.

We can check by taking the derivative using the slope function at a point, say x = 1.  The derivative of that is 23.

Now use the mean value theorem, [f(b) - f(a)]/(b - a).  Use values for a and b very close to 0, one on either side to approximate the derivative at 0.

a =1.1
b =0.9

f(0.9) = 11.846
f(1.1) = 16.454

Therefore, we get (11.846 - 16.454)/(0.9 - 1.1)  = 23.04  so the slope function holds true.

Here's some simple rules to remember for logarithms.

log(base b) x = y can be rewritten in exponential form as b^y = x.

 Ex)  log (base 3)9 = 2 is the same as 3^2 = 9

ln x = y is the same as log (base e) x = y, which is e^y = x.

A natural log is basically a logarithm with base e, where e is approximately 2.718.

The log of a product is the same as the sum of logs.

log xyz = log x + log y + log z

The log of a quotient is the difference of the logs.

log x/y = log x - log y


Another useful rule is log x^2 = 2logx.  Basically the exponent can be moved in front of the log.

These basic rules will help anyone when trying to solve problems with logarithms.

Suppose a function  f(x) is differentiable and continuous on the interval [a,b]. If there exists a number c in the interval [a,b] such that f '(c) = 0, then the Mean Value Theorem applies.

For example..  suppose f(x) = x^2 + 3x. We want to test the Mean Value Theorem over the interval [-2, 1].

f'(x) = 2x + 3

Now set f'(x) = 0 and solve for x.

2x + 3 = 0, therefore x = -3/2.  Since -3/2 falls in the interval, the Mean Value Theorem applies.

An example where it doesn't apply. Suppose the interval is [-5,5]

f(x) = 2x..

f'(x) = 2

there is no value c where the derivative is 0 since the derivative everywhere along the function is 2. Therefore, the Mean Value Theorem does not apply.

Suppose you want to drain a swimming pool at the end of the summer and you know the pool drains at 10 cubic feet per minute. The pool is 100 feet long and 40 feet wide. For the first 50 feet, the pool was water depth of 4 ft. For the next 30 feet the water depth is 7 feet and for the final 20 feet the water depth is 10 feet. How long will it take the pool to drain?

We need to calculate the volume of the pool.  Recall the volume is length times width times depth.

For the first 50 feet the volume is 50(40)(5) = 10,000 cubic feet.  For the next 30 feet the volume is 30(40)(7) = 8,400 cubic feet. For the last 20 feet, the volume is 20(40)(10) = 8,000 cubic feet. Therefore the total volume of the pool is 26,400 cubic feet.

The time it takes the pool to drain is 26,400/10 = 2,640 minutes.