The max's and min's a function on a closed interval can occur either at local extrema, or the endpoints. Local extrema occur when the derivative is 0. First, taking the derivative,

We see that the extrema will occur at  and. Here, we can either check to see which of these extrema are max's using the first or second derivative test, or we can just plug them into our function -- if one of them was actually a min, it just won't be our answer. 

If you choose to go the first route, you have 

So we know that the extrema at 0 is a local max, and at 2/3 is a local min.

Testing our max and two endpoints, we have

So our maximum value is 7.

Let the dimensions of the picture be x and y.

Our constraint is

Solve for x

Let us substitute that value of x back into our area equation

In order to maximize this equation, we need to find its derivative and set it equal to zero

Now that we have found our first dimension, we need to find the second

Then our dimension are 10in and 20in

Points of inflection are found when the second derivative of a function is equal to zero. For our function

The first derivative is:

By the product rule.

Our second derivative then is:

Setting this equal to zero we obtain:

Since we know  can never be equal to zero, we are only concerned with , so 

Maximum's and minimums of functions can take place where the derivative is 0 or undefined. To find out where this occurs, we take the derivative.

As you can see, the derivative will be 0 at .

At this point, we have three options:

We can plug them all into the original equation and see which are the biggest and which are the smallest.

We can use the first derivative interval test to see which are mins and which are maxes.

We can use the second derivative test to see which are mins and which are maxes.

As we're likely going to have to plug them all into the equation anyway, and the problem doesn't ask to classify the points as local mins or maxes, the first approach is the most time effective. Plugging in our points we find that  ;  ;

Before we determine the answer, it's important to remember that the global maxes and mins on a closed domain can occur at the endpoints, and on an open domain, may not exist. For this problem, we're given a quartic that faces upwards. Thus, there is no global maximum because the graph shoots off to infinity on either side. On the other hand, there is a global min because the graph is continuous and does not go down to negative infinity.

As we know the min must occur at one of the points we've determined above, we can see that we have a global min of -123 at x = 4 and no global maximum.

Also note, as a bit of terminology, that a global maximum or minimum value is a value that occurs at a point. Thus our answer isn't that the maximum value is (4, -123), but just -123.

This question is an optimization problem. This question asks to find the distance to run along the beach that minimizes the time it takes to get to the swimmer in the water. To minimize time, we need to construct an equation, where time is a function of one variable. Fortunately, the equation

can be solved for time, and this is how we will create the equation we need.

Solving this equation for time gives

We have two Rates. Running on the beach, and Swimming in the water. So we have two times to consider:

and

Adding these two times gives the total time to get to the person in the water.

We are given the rates that the life guard can run and swim, 10ft/sec and 4ft/sec respectively. The variable is how far to swim down the beach.

Labeling the distance the life guard runs as will make the math somewhat difficult, where as labeling the distance the life guard runs as will make the math a little nicer. We just need to remember that we did this. The picture below shows the labeling used in this explanation.

With this labeling, the distance the life guard runs is , and the distance the life guard swims is by Pythagorean Theorem.

Plugging the distances and rates into our Time equation gives:

This expresses Time as a function of one variable, . This is what we need to minimize. To do this we will find the relative minimum of this function. So we find the first derivative. First, we will do a little algebra and split the into two fractions and rewrite the square root as an exponent to make the derivative easier to compute.

Now we find the derivative. The derivative of is .

The derivative of  is .

We use the chain rule for . Doing so gives

Assembling the pieces results in the following derivative

To find the critical points, we substitute 0 in for Time'.

To solve for , we move the  to the opposite side and then cross multiply.

 Now we can divide by 4 on both sides to isolate the square root, then reduce the resulting 10/4 to 5/2. Then we square both sides to eliminate the square root.

We know need to move the terms to the same side, the right side in this explanation Remember that we will need to get the common denominator to combine them.

Multiply both sides by to isolate . Then square root both sides. We will not need to incorporate the when we square root, since  is a physical distance.

Since the question asks us to approximate the answer to the nearest hundredth, we can plug into a calculator to get a decimal. Doing so gives approximately x=43.64 ft.

However, this is not our final answer. Recall that we defined the distance the life guard ran as . So we need to subtract x=43.64 from 200 to find what the question is asking for. 

Doing this we get our final answer to be .

Since the rectangle has one corner on the origin, (0,0), and the opposite corner at a point (x,y) on the graph, we can set the width = x, and the length = y. 

The area of a rectangle is , so for this problem, .

Since we are trying to maximize the area, we need to express area as a function of only one variable. Right now we have two variables. The way to fix this is to subsitute one variable, like , with an equivalent function of .

Most problems give a relationship between the two variables. In this problem, the equation of the quarter-ellipse, , is the relationship we need.

We can substitute the in the Area equation with what is equals.

Now we have the equation of what we want to maximize, Area , written in terms of only one variable, x. Now we can find the relative maximum by finding the derivative.

Before finding the derivative, it will be helpful to rewrite the square root as an exponent.

We will need to use the product rule, , and use chain rule for . Doing this, we get:

Now we find the critical points of the derivative by setting it equal to zero and solving. We will simplify the derivative slightly at the same time. We can multiply the fractions in the second part, an move the  down to the denominator to remove the negative from the exponent. This gives us:

Now we can solve for x. First, subtract the term to the left side.

Now multiply both sides by denominator, , in order to eliminate the fraction.

The is multiplied by the same group, so we add their exponents, 1/2 + 1/2, which is 1. So we don't need to write the new exponent, since 1 is understood.

Since the power on the group is 1, we can distribute the -6 through it.

Moving all the terms to the left, and the -6 to the right, we get:

Divide both sides by 2 to isolate

Now square root both sides.

Since the graphs domain is given as , we can ignore the negative answer and use the positive answer.

Now we know the width is . To find the length, we need to find . 

Plug into the graph's equation to find .

We need to get the square root out of the denominator. To do this we mulitply the numerator and denominator by .

Thus, the length of the rectangle should be .

Now we know the dimensions of the rectangle, that maximize its area:

width=

length =

To find the coordinates of the local extrema of a function, we need to find the critical points of its first derivative.

Since is a polynomial, we can find its derivative term by term. The first 3 terms can be differentiated using the power rule, , and the constant multiple rule, .

The last term is a constant, and its derivative is zero.

Applying these rules, we find the first derivative:

Now we need to find the critical points. To do this, we set the first derivative equal to zero and solve. Factoring is the best method in this problem.

Now that we have the critical points, we need to determine for each one, whether it is a maximum, minimum, or neither. We use the first derivative line test to determine this.

For , we will test the interval before it, , and the interval after it, , and find whether they are increasing or decreasing.

For the interval , we will test and find whether is positive or negative.

Since is positive, the original function is increasing before the critical point, .

Now we will test the interval after .

For the interval,  , we will test and find whether is positive or negative.

Since is negative, the original function is decreasing in the interval following .

Since the function is increasing before , and decreasing afterward, we can conclude that a maximum occurs at .

Now we find the value of this maximum, .

Thus  is a local maximum.

Now we will determine whether a maximum or minimum occurs at .

We know that is decreasing before , but we still need to determine what happens afterward.

For the interval , we will test , and find whether is positive or negative.

Since is positive, the original function is increasing following .

Since is decreasing before, and increasing after , we can conclude that a minimum occurs at .

Now we need to find the value of this minimum, .

Thus is a local minimum.

So our answer is:

is a local maximum.

is a local minimum.

Concavity refers to the "curving" of the function. While the first derivative describes when the function is increasing or decreasing (instantaneous rate of change of ), the second derivative describes concavity, the instantaneous rate of change of .

The first derivative is like velocity, (moving forward or backward),  while the second derivative is like acceleration (speeding up or slowing down).

Since we need to find the intervals of concavity, we will find the second derivative and work with it.

First we must find the first derivative using the power rule and constant multiple rule for each term of .

This gives:

Now to find the second derivative, we take the derivative of .

The same derivative rules apply:

Now that we have the second derivative, , we must find its critical points. We do this by setting , and solving.

The best method for this case is factoring.

The greatest common factor is ,

Inside the parentheses is a quadratic expression that can be factored like so:

Setting each factor equal to zero we find the following:

Now we know the critical points for the second derivative. To find the intervals of concavity, we test a point in each interval around these critical points and find whether the second derivative is positive or negative in that interval.

For the interval , we can test .

Using the factored form of the second derivative is easier than using the polynomial form, since the arithmetic involves fewer large numbers.

Since is negative, the original function is Concave Down in the interval .

Now for the interval , we can test .

We will combine the fractions inside the parentheses by getting the common denominator.

Multiplying the fractions gives:

Since is positive, the original function is Concave Up on the interval .

For the interval , we can test .

Since is negative, the original function is Concave Down on the interval .

Finally, for the interval , we can test .

Since is positive, the original function is Concave Up on the interval .

Summarizing the results, the intervals of concavity are:

Concave Down: 

Concave Up:

To find if the equation is increasing or decreasing, we need to look at the first derivative. If our result is positive at , then the function is increasing. If it is negative, then the function is decreasing.

To find the first derivative for this problem, we can use the power rule. The power rule states that we lower the exponent of each of the variables by one and multiply by that original exponent.

Remember that anything to the zero power is one.

Plug in our given value.

Is it positive? Yes. Then it is increasing.

To find the concavity, we need to look at the second derivative. If it is positive, then the function is concave up. If it is negative, then the function is concave down.

Repeat the process we used for the first derivative, but use  as our expression.

For this problem, we're going to say that  since, as stated before, anything to the zero power is one.

Notice that  as anything times zero is zero.

As you can see, there is no place for a variable here. It doesn't matter what point we look at, the answer will always be positive. Therefore this graph is always concave up.

This means that at our given point, the graph is increasing and concave up.

To answer this question, one first needs to find  and then find the critical points of the function (i.e. where .  Finally, one would need to determine the sign of  for the intervals between the critical points.  

For the given function:

 .

Therefore,  when  and .  So, the intervals to consider are:

To determine the sign of , pick any number for the given interval and evaluate  at that number.

Therefore,  is increasing on the intervals  and  since  is greater than zero on these intervals.