We are differentiating  with respect to . So we can treat  as a constant.

Therefore, the derivative becomes .

By plugging in   and , we get .

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 dimensions are  inches and  inches.

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  at  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 .

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 lifeguard can run and swim, 10ft/sec, and 4ft/sec respectively. The variable is how far to swim down the beach.

Labeling the distance the lifeguard runs as  will make the math somewhat difficult, whereas labeling the distance the lifeguard 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 the area as a function of only one variable. Right now we have two variables. The way to fix this is to substitute 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 domain of the graph is given as , we can ignore the negative answer and use a 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 multiply 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 =

The derivative of  is .  So the graph of the derivative will have a positive slope of two and pass through the origin.

The local maximums will be the points at which the slope is equal to zero and the slope is in the process of changing from positive to negative.  The local minimums will be the points at which the slope is equal to zero and in the process of changing from negative to positive.  Therefore the maximums are  and the minimums are  .

Points of inflection occur when the graph of the derivative cross the x-axis. If the derivative is going from positive to negative, this tells us the function has a local maximum at this point.  If the derivative is going from negative to positive, then the function has a local minimum at that point.  From this we see that there is a local maximum at  and a local minimum at .

The function is increasing when its derivative is positive and decreasing when its derivative is negative.  From this we see that the function is increasing on the intervals and  and decreasing on  .

The first derivative graph can also tell us about the concavity of our function.  When the derivative is increasing, the function is concave up.  When the derivative is decreasing, the function is concave down.  From this we see that the function is concave up on the interval  and concave down on .