Maxima/Minima Problems

[drbeat14]

Calc 2 is discussing maxima and minima, and I thought I understood, but I cannot figure out what exactly is being asked to find, and what step 1) would be for two problems on my practice sheet. I have attached the problems, they are #26, #27. Any help would be appreciated.

 

[Quaid]

Hi drbeat14:

Your image is difficult to read, even when viewing the enlarged version. In the future, please attach one image for each exercise. Also, the board guidelines request that you start a separate thread for each exercise. Please keep this in mind, for the future.

I'll help you with (27); I'm not sure why you posted (28).

There are many different rectangles that can be formed; the rectangle they drew is one possibility. (I added two more example rectangles, in the image below.)

Clearly, there is an (x,y) point on the curve at the upper-right corner of every possible rectangle in Quadrant I.

They want you to find the specific point (x,y) which results in the rectangle with the largest area.

The area of each rectangle is WIDTH times HEIGHT.

The width of each rectangle is the same as the x-coordinate of the point at the upper-right corner. Is that clear?

The height of each rectangle is the same as the y-coordinate of the same point. Okay?

Hence, the area of these rectangles is x*y.

To write the area function in terms of x only, you need to express y in terms of x.

So, the first step is to solve the given equation for y. Please show us what you get.

Once you have the area function in terms of x only, you find the maximizing x-value through the usual process of (1) determining the function's first derivative and (2) solving for the critical point.

If you're still stuck, please reply with specific explanations or questions about what you've tried so far.

Cheers :cool:

(Click thumbnail for larger image)

 

[JeffM]

Calc 2 is discussing maxima and minima, and I thought I understood, but I cannot figure out what exactly is being asked to find, and what step 1) would be for two problems on my practice sheet. I have attached the problems, they are #26, #27. Any help would be appreciated.


View attachment 3691

Are you asked to solve problem 27 in any particular way? You seem to be studying multivariate calculus.

Simple way to do problem 27 is, as was mentioned in the preceding post, to turn it into a problem in one variable, but it can be solved in multiple variables using the method of Lagrangian multipliers. How would you set that up?

 

[drbeat14]

Critical Point?

Thank you for responding. I am sorry that entire page got scanned, I completely forgot about cropping and enlarging it. The equation I came to is this, though I am not certain if it is correct thus far, and if so, how to find the critical value. I attempted to type in word this time, I apologize, as I am still adjusting to the math symbol system on here. Let me know if this seemed more clear, or worse. Thanks again.

 

[drbeat14]

PS-

Also thank you Quaid for enlarging that...it's easier for me to see as well now. Jeff, it does not say we should show any particular method. Thx.

 

[Quaid]

The equation I came to is this ...

-x/[4*sqrt(4-x^2/4)] = 0

Is that your equation for finding the critical point? I asked you to show me your result for solving the given equation for y.


The area function that I came up with is

x/2*sqrt(16 - x^2]

 

[drbeat14]

Re:Quaid

No, I was attempting to do what you asked, I just think I did it incorrectly.

 

[Quaid]

x^2/16 + y^2/4 = 1

To solve the given equation for y, we need to end up with something that looks like this:

y = [an expression containing only x]

In other words, you need to isolate y on one side of the equals sign; everything else goes on the other side.

To isolate y, begin by isolating the y^2/4 term. (Subtract x^2/16 from both sides...)

Can you continue?

 

[drbeat14]

This is what I have thus far, let me see if I can make it a clearer image on the next one...

 

[drbeat14]

Ok, I am now getting the same equation Quaid got, except I keep getting it in negative:

- x/ (2 sq(16-x^2)

 

[Quaid]

That's good. (There ought to be a plus-or-minus sign, in front of the square root; hopefully, you left it off because you realized that we're dealing with the curve in Quadrant I only).

We can simplify that radical. Factoring out 1/16 from (1 - x^2/16) gives us

sqrt([4/16]*[16 - x^2])

This simplifies to y = 1/2*sqrt(16 - x^2)

Did you follow that? :cool:

 

[Quaid]

now getting the same equation Quaid got, except I keep getting it in negative:

- x/ (2 sq(16-x^2)

We just cross-posted.

The radical should not be in the denominator; or, is that just a typographical error? (Otherwise, you're missing a closing parenthesis...)

Also, if this is the expression for y, then there should be no x in the numerator.

 

[drbeat14]

I think it shows up in my email before in the post. Anyway yes it is a typo...

 

[Quaid]

That's okay; these are issues that arise when people are conversing by thread posts in real-time.


So, your previous post is not showing the expression for y that I asked you to show.

This is your definition for the area function, correct?

f(x) = x/2 sqrt(16 - x^2)


(I cannot determine how you obtained the leading negative sign, unless you show your work.)


If you're ready to continue, please show your work on the derivative.

 

[drbeat14]

You know, I think that I do not have it written correctly, since the format is confusing me, attached is handwritten of what I have, I don't have a clue why it is sideways, btw

 

[drbeat14]

Followup-Q.

Ok, I know my answer has to be b) or c), but I think I shall have to consult the textbook on critical values, as I still do not know where to go from the derivative equation. I am going to take a break from this for the evening. Thank you for all of your help. Please let me know if you have any music questions I can help with to return the favor...a much less useful subject, I know.

 

[Quaid]

I do not know why the image is rotated; maybe it's a camera issue, maybe it's a software issue. Our boards display image files exactly as they are uploaded. You'll need to resolve issues with image files before attaching them. (If you use Windoze, the MSPaint application can rotate an image and resave the file.)

The radical should not be in the denominator. You still have not shown any work, so I cannot determine how you ended up writing it that way.


y = 1/2 * sqrt(16 - x^2)

The area of a rectangle is x*y

Substituting the expression for y gives the area of a rectangle as

x * 1/2 * sqrt(16 - x^2)


Therefore, the area function is

f(x) = x/2 * sqrt(16 - x^2)


What would you like help with next? Do you want to clear up your confusion about getting the correct f(x)?

:cool:

 

[Quaid]

I shall have to consult the textbook on critical values, as I still do not know where to go from the derivative equation.

That's okay.

When you're ready to continue, please explain where you're at in the exercise (we did not get to the derivative of function f, yet).

Cheers :cool:

 

[drbeat14]

Thanks to Quaid

Hi there

Thanks for your help and being patient. I reviewed the critical values and it simply said to set the derivative function to zero. Which I did and concluded that answer b must be correct. However, to be honest I used a derivative calculator. So at this point obviously my problem is that I am not doing derivations correctly because my result looked absolutely nothing like what was displayed as the correct one. When you get a chance, would you mind getting me started on the first few steps of the d/dx so I can try it and then upload that much and see if it looks right to you?

 

[JeffM]

Hi there

Thanks for your help and being patient. I reviewed the critical values and it simply said to set the derivative function to zero. Which I did and concluded that answer b must be correct. However, to be honest I used a derivative calculator. So at this point obviously my problem is that I am not doing derivations correctly because my result looked absolutely nothing like what was displayed as the correct one. When you get a chance, would you mind getting me started on the first few steps of the d/dx so I can try it and then upload that much and see if it looks right to you?

I am going to go all the way back to the beginning.

Name your variables

area of rectangle = xy. What is to be maximized is a.

Write down your constraints: $\displaystyle \dfrac{x^2}{16} + \dfrac{y^2}{4} = 1,\ x \ge 0, y \ge 0.$

$\displaystyle But\ y \ge 0 \implies y = 0.5\sqrt{16 - x^2} \implies A = 0.5x\sqrt{16 - x^2}.$ You eventually got here.

So far this is all preliminary algebra.

Now I would use substitution, the multiplication rule, the power rule, and the chain rule. I am aiming for something like a = uw that's simple to work with.

$\displaystyle u = 0.5x \implies \dfrac{du}{dx} = what?$ First substitution and multiplication rule.

$\displaystyle v = 16 - x^2 \implies \dfrac{dv}{dx} = what?$ Second substitution and power rule.

$\displaystyle w = \sqrt{v} = v^{(1/2)} \implies \dfrac{dw}{dv} = what? \implies \dfrac{dw}{dx} = what?$

Third substitution, power rule, and chain rule

$\displaystyle And\ a = uw \implies \dfrac{da}{dx} = what?$ Multiplication and chain rules

You can do all this in one fell swoop without any substitutions, but, until you get experience, I suggest breaking it up into pieces like this. It is a bit complex to figure out what the pieces should be, but each one should be easy to work with.

Once you have $\displaystyle \dfrac{da}{dx}$, set it equal to zero and solve for x, remembering that $\displaystyle x \ge 0$.

Once you know x, you can compute y.

Give it a try