Melody

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Melody  Feb 11, 2022
 #8
avatar+118723 
0

Mon 26/10/15

WOW   The web2.0 calc is back on the forum.      Hiphip horray!!!!       Thanks Admin!!

 

1)  How to integrate k^sin(x)

     http://web2.0calc.com/questions/how-to-integrate-a-periodic-exponent-function#r1

 

2) This is a great example of mathematical fallacy. :)

    http://web2.0calc.com/questions/mathematical-fallacy

 

3) Median age (for younger students)

   http://web2.0calc.com/questions/what-is-the-median_2#r1

 

4) Velocity, acceleration and rotational movement.   Thanks Alan.

    http://web2.0calc.com/questions/curve-motion

 

5)  How steep are the spiral stairs?     Thanks Rom and Melody

     http://web2.0calc.com/questions/sin-cos-tan_3#r2

 

6)  How many multiple choice questions are there?     Melody

   http://web2.0calc.com/questions/please-help_67706#r1

 

7)  FUN with logic       Thanks guest     cheeky

    http://web2.0calc.com/questions/math-question_23564#r2

 

8)  System of equations    Thansk guest

     http://web2.0calc.com/questions/solve-the-following-system-of-linear-equations#r2

 

9) Who knew that pizzas could be so confusing!

    http://web2.0calc.com/questions/pizza#r1

 

10)  Problems when taking fractional indices of negative numbers

    http://web2.0calc.com/questions/wonky-calculator#r5

Oct 25, 2015
 #10
avatar+118723 
+21

@@ What is Happening?  [Wrap4]    Sun 25/10/15   Sydney, Australia Time 8:09pm   ♪ ♫

 

Hi everyone,

 

There have been a couple of interesting posts today I think.  smiley

 

1)  Trig manipulation question       Thanks Omi67 and Melody.

      http://web2.0calc.com/questions/trig-help_4130

 

2) Maximizing a function given a system of contraints           Melody

    http://web2.0calc.com/questions/linear-programming-pls-help#r1

 

3) I got a bit carried away with this differentiation.  ://      Thanks also to Omi67

    http://web2.0calc.com/questions/differentiation_1#r1

 

4) Trig, find the height of the helicopter.    Thanks Alan.

     http://web2.0calc.com/questions/trigo-sum

 

 

If you want to comment an any issues please do so.    

 

Enjoy the rest of your day :) 

 

Melody    ♪ ♫

 

----------------------------------------------------

 

Lantern thread

Oct 25, 2015
 #2
avatar+118723 
+10

PayPay sent me a private message asking if i could try to explain some more.  

Paypay doesn't understand the moving orange line.  

It is a strange concept I think, I am not surprised that you did not understand and I am pleased that you asked about it. laugh

 

 

I have changed the graph just a little. (I just added the names of the points)

https://www.desmos.com/calculator/4w3xa3c3yy

 

x+y≤8

2x+y≤10

x≥0

y≥0

 

Now the dark quadrilateral in the middle is the region that x and y must lie in.

The vertices of this region are  (0,0),     (5,0),     (0,8),      (2,6)

Lets just substitute these points into the equation  100x+40y = constant

(0,0)       100*0+40*0=0                           The constant would be 0

(5,0)         100*5+40*0=500                     The constant would be 500

(0,8)        100*0+40*8=320                      The constant would be 320

(2,6)        100*2+40*6=200+240=440    The constant would b e 440

 

So the biggest constant is 500.   It happens whe x=5 and y=0     (5,0)

So you can do this without understanding the orange line but it would be good if you understood the line so I will try to explain.

 

 

Now, what has this got to do with the moving orange graph???

 

The orange line is the graph of  100x+40y = a constant 

What you are finding is the biggest constant that  this equation can equal (within the given region.)

The slider is N

So you can make N bigger or smaller.   I have set it to move between 0 an 1000.

So you can move the orange line, by changingthe value of N.   You want the biggest possible N value so that at least some part of the line falls in the given region. When N is 500 only one point of the line is in the region.  That point is (5,0).   If N is bigger than 500 then no point on the line will fall in the restricted region.  SO the biggest possible value of N within the given restrictions is 500 and N is 500 at the point (5,0)

 

Have a good think about this because it is a concept that can help you with a variety of problems.  :)

Oct 25, 2015
 #1
avatar+118723 
+5

\(f(x)=\frac{x^3+2 cos(x)}{2sin(x)}\\ f'(x)=\frac{(2sinx)(3x^2-2sin(x))-2cos(x)(x^3-2cos(x))}{4sin^2(x)}\\ f'(x)=\frac{(2sinx)(3x^2-2sin(x))}{4sin^2(x)}-\frac{2cos(x)(x^3-2cos(x))}{4sin^2(x)}\\ f'(x)=\frac{(3x^2-2sin(x))}{2sin(x)}-\frac{cos(x)(x^3-2cos(x))}{2sin^2(x)}\\ f'(x)=\frac{3x^2}{2sin(x)}-1-\frac{x^3cos(x)-2cos^2(x)}{2sin^2(x)}\\ f'(x)=\frac{3x^2}{2sin(x)}-1-\frac{x^3cos(x)}{2sin^2(x)}-\frac{cos^2(x)}{sin^2(x)}\\ f'(x)=\frac{3x^2}{2sin(x)}-1-\frac{x^3cos(x)}{2sin^2(x)}-(tan(x))^{-2}\\ \)

 

\(f '(\pi/2) = \frac{3*\frac{\pi^2}{4}}{2}-1-\frac{\frac{\pi^3}{2^3}*0}{2}-0\\ f '(\pi/2) = \frac{3\pi^2}{8}-1\\ f '(\frac{\pi}{2}) = \frac{3\pi^2-8}{8}\)

 

 

 

\(\qquad\frac{d}{dx}\;\frac{x^3cos(x)}{2sin^2(x)}\\ \qquad=\frac{(2sin^2(x))*[3x^2cos(x)-x^3sin(x)]-4sin(x)cos(x)*x^3cos(x)}{4sin^2(x)}\\ \qquad=\frac{[6x^2sin^2(x)cos(x)-2x^3sin^3(x)]-4x^3sin(x)cos^2(x)}{4sin^2(x)}\\ \qquad=\frac{3x^2sin(x)cos(x)-x^3sin^2(x)-2x^3cos^2(x)}{2sin(x)}\\ \qquad=\frac{3x^2sin(x)cos(x)}{2sin(x)}-\frac{x^3sin^2(x)}{2sin(x)}-\frac{2x^3cos^2(x)}{2sin(x)}\\ \qquad=\frac{3x^2cos(x)}{2}-\frac{x^3sin(x)}{2}-\frac{x^3cos^2(x)}{sin(x)}\\ \)

 

\(f'(x)=\frac{3x^2}{2sin(x)}-1-\frac{x^3cos(x)}{2sin^2(x)}-(tan(x))^{-2}\\ f''(x)=\frac{12xsin(x)-6x^2cos(x)}{4sin^2(x)}-\left[\frac{3x^2cos(x)}{2}-\frac{x^3sin(x)}{2}-\frac{x^3cos^2(x)}{sin(x)}\right]+2(tan(x))^{-3}(sec(x))^2\\ f''(x)=\frac{12xsin(x)-6x^2cos(x)}{4sin^2(x)}-\left[\frac{3x^2cos(x)}{2}-\frac{x^3sin(x)}{2}-\frac{x^3cos^2(x)}{sin(x)}\right]+\frac{2cos^3(x)}{sin^3(x)cos^2(x)}\\ f''(x)=\frac{12xsin(x)-6x^2cos(x)}{4sin^2(x)}-\frac{3x^2cos(x)}{2}+\frac{x^3sin(x)}{2}+\frac{x^3cos^2(x)}{sin(x)}+\frac{2cos(x)}{sin^3(x)}\\ f''(\frac{\pi}{2})=\frac{6\pi *1-0}{4*1}-\frac{3x^2*0}{2}+\frac{\pi^3}{16}+\frac{x^3*0}{1}+\frac{0}{1}\\ f''(\frac{\pi}{2})=\frac{3\pi}{2}+\frac{\pi^3}{16}\\ \)

 

Oh dear, I thought I was supposed to find f ''  Maybe I was only supposed to find f '         blush

oh well I will go back and do that too.     angel

.
Oct 25, 2015
 #1
avatar+118723 
+15

Hi PayPay      laugh

 

By graphing the systems of constraints, find the values of x and y that maximize the objective function.

 

x+y≤8

2x+y≤10

x≥0

y≥0

 

Here is your graph.   smiley  

https://www.desmos.com/calculator/lmiz3pkgw2

 

The constraints for a quadrilateral where the vertices are the 4 points that you have been given

 

I have also graphed 100x+40y=N  where N can take on a sliding range of values.

It is orange and if you move the slider the orange line will move.

The greatest value of N will occur on the last corner befor the orange line leaves the constraint region.  Hopefully you can see that happen at (5,0)

maximum for N = 100x+40y 

 

Alternatively, if you substitute those answer values into the N equation, you will find that N is greatest at (5,0)

 

answers:

(0,0)

(5,0)

(0,8)

(2,6)

Oct 25, 2015