Currently taking Cal 1. I have an exam this Friday and still not strong solving the following problems from my book section. Can any one please show me the steps and the answer.

-Find the absolute extreme values of the function on the interval.

f(x) = tan x, -π/4 ≤ x ≤ π/6

Just in case you where wondering the following symbol is PI: π


-Determine the dimensions of the rectangle of largest area that can be inscribed in a semicircle of radius 3.


Thank you so much!:o

1. I'm not sure, but I'm not familiar with this wording... does extreme values include inflexion points too?

Because the graph of f(x) = tan(x) doesn't have any maxima or minima...

or I'm overlooking something.

2. Make a sketch.

Let's say the length of the rectangle is l and the width w.

Area of rectangle = lw

Draw a line from the centre of the semicircle to the vertex of the rectangle. This is the radius, right?

We get:

3^2 = w^2 + \(\frac{l}{2}\)^2

36 = 4w^2 + l^2

Use that in the above equation, and we get:

Area = (w)(\sqrt{36 - 4w^2})

Now, find the derivative of Area with respect to w, and set this to 0 and solve for w to get the width when the area is a maximum.

Thank you for the rapid response.

does extreme values include inflexion points too?
YES!

When solving this the problem came because from what I learned in trigonometric functions tan line does not have min max. However, my professor (after I clearly emailed her 20 times) answered that a problem very similar to this will be in my final tomorrow. I am good with derivatives and l'hopital and intervals, etc. This however I just can't put my mind to rest.

Thank you again.

As for the second problem. I am going to finish the calculations at the end and get back to you, so I can compare.

If you have any more information in the first problem please explain.
Thank you again

Ah! That settles it then :)

Find f'(x).

\frac{d}{dx}(\tan(x)) = \sec^2(x) =\tan^2(x) + 1

Set this to zero.

\tan^2(x) + 1 = 0

As you can see, there are no solutions, hence no extreme values.

(I forgot something earlier)

If you draw the graph of tan, you know that there are no maxima and minima. You might perhaps think about when theta = 0, but then, the gradient is tan^2(0) + 1 = 1 which proves that it's not an inflexion point :)

AHH I see your point. Let me see if I can solve. I will post wheni have done.

Thank you for all the help

Also, if it is not so much to ask. I have done problems which I am ot sure I did right and I don't have the answer key. Can you please check them for me?


A company is constructing an open-top, square-based, rectangular metal tank that will
have a volume of 60 ft3. What dimensions yield the minimum surface area? Round to the
nearest tenth, if necessary.

answer
Let the dimensions of base of tank = x ft

and height = h ft

Volume= x2h ft3= 60

Area of tank= x2 + 4xh

Expressing area in single variable,

A= x2 + 4x(60/x2)

A= x2 + 240/x

For minium area , A'=0

A'= 2x =0

2x = 240/x2

x3 = 120

x = 4.93 ft approx ( dimensions of length and width)

and height = 2.47 ft approx

Well, it's good though I don't agree with this line:


A'= 2x =0

To be more explicit, write it all

A' = 2x - \frac{240}{x^2}

Or if it was a typo, it's all right :)

Oh it was a typo. I had it done in word and then when copying it here it did not go trhough

Thank you

For the question about extremes - on the interval \frac {-\pi} {4} \le x \le \frac {\pi} 6 there are indeed maximum and minimum values for f(x) = \tan(x) . If you plot the tan(x) function you'll see that the function increases monotonically in this domain - therefore it's maximum in this interval is at x = \frac {\pi} 6 .

Does extreme values here also included the points where the gradient is not zero?

If so, then there would be two extreme values. One at pi/6 and another at -pi/4