Maximizing Area: Calculating Triangle and Rectangle Dimensions | Homework Help

[joemama69]

Homework Statement

Note the picture

Homework Equations

The Attempt at a Solution

A of both trianles = xsinQ\sqrt{x^2 - x^2(sinQ})^2

A of rectangle = (w-2x)(xsinQ)

Total A = (w-2x)(xsinQ) + xsinQ\sqrt{x^2 - x^2(sinQ)^2}

A = wxsinQ + 2x²sinQ + x²sinQcosQ

A_x = wsinQ + 4xsinQ + 2xsinQcosQ = 0

-wsinQ = x(4sinQ + 2sinQcosQ)

x = -w/(4 + 2cosQ)

A_Q = wxcosQ + 2x²cosQ + x²(sinQ)² - x²(cosQ)² = 0

It gets pretty messy when i plug in x, are there anny errors so far

 

[joemama69]

anybody, can i get some help

 

[Mark44]

Homework Statement

Note the picture

Homework Equations

The Attempt at a Solution

A of both trianles = xsinQ\sqrt{x^2 - x^2(sinQ})^2

A of rectangle = (w-2x)(xsinQ)

Total A = (w-2x)(xsinQ) + xsinQ\sqrt{x^2 - x^2(sinQ)^2}

A = wxsinQ + 2x²sinQ + x²sinQcosQ

A_x = wsinQ + 4xsinQ + 2xsinQcosQ = 0

-wsinQ = x(4sinQ + 2sinQcosQ)

x = -w/(4 + 2cosQ)

A_Q = wxcosQ + 2x²cosQ + x²(sinQ)² - x²(cosQ)² = 0

It gets pretty messy when i plug in x, are there anny errors so far

What's the problem you are trying to solve? The picture doesn't give me any idea of what you're trying to do.

 

[joemama69]

you must Maximize the area, note the title

 

[Mark44]

Some context would have been nice, such as that the area is the cross-sectional area of what looks to be a trough, formed from sheet metal that is W units wide.

Your picture doesn't convey that information, which would have been helpful.

You have a sign error in this line:

A = wxsinQ + 2x²sinQ + x²sinQcosQ

The first '+' sign should be '-'. That makes A_x = WxsinQ - 4xsinQ + 2xsinQcosQ = sinQ[W -4x + 2xcosQ].

The sign error also affects A_Q, which I get as xWcosQ -2x²cosQ + x²cos²Q - x²sin²Q.
About the only thing I can think of to do is to factor x out of each term in this partial.

The critical points are where Q = 0 or when x = 0, either of which gives you an area of 0, and whatever you get from some messy equations that remain. One of these is W - 4x + 2xcosQ = 0, which you can solve for x in terms of Q (and W), or solve for Q in terms of x (and W). Whichever variable you solve for, substitute that in the other equation that comes from A_Q.

 

[joemama69]

Ya, this is pretty much where i ran into problems, i pluged in my x = w/(4-2cosQ) into A sub Q

A_Q = 0 = (w²cos)/(4-2cos) - 2w²cos/(4-2cos)² + (2w²cos²)/(4-2cos)² - (w²sin²)/(4-2cos²)

I would have to solve that for Q (i got lazy and didnt type them, just assume they are there), which means (tell me if I am wrong), i only have to solve the numerators of the expression for 0 in terms of w

I got Q = 0 & 2pi which would make the gutter flat, what happened

 

[Mark44]

I didn't work it through that far, so I'll assume that the work you have in post 6 is correct. You can't just "solve the numerators," as you put it, since not all of your expressions have the same denominator; the first has a denominator of 4 - 2cosQ. You'll need to multiply that term by (4 - 2cosQ)/(4 - 2cosQ), and then you can factor out 1/(4 - 2cosQ)² from all four terms. Finally, you'll need to see what values of Q (theta) make A_Q equal to 0.

As I said before, Q = 0 and x = 0 are critical values, but they produce a trough with cross-sectional area 0, so they are not the values you want.

 

[joemama69]

ok i was able to factor out the w^{2} and the (4-2cos)^{2} and I got

4cos - 2cos^{2} - 2cos + cos^{2} - sin^{2} = 2cos - con^{2} - sin^{2} = 0


I got Q = pi/3, 5pi/3, 7pi/3

how do i calc which one is the maximum, there are many possibilities

 

[Mark44]

Due to the fact that you're working with a physical object, a trough, there is a natural range of values for Q (theta). That should help you eliminate many of the possibilities. If A_xx < 0 and A_QQ < 0, you're at a local maximum.