Recent content by IsaacStats

Hello everyone, I have the following question: Show without using the simplex method that x1=5/26, x2=5/2, x3=27/26 is an optimal solution to the following LPP. Maximize z=9x1+14x2+7x3 subject to 2x1+x2+3x3<= 6 5x1+4x2+x3<= 12 12x2 <= 5 x1,x2,x3 unrestricted...

EXCELLENT! THANKS! I believe my mistake was looking at 0<=rsin(theta)<=1. Thank You for your help and may numbers be in your favor.

Out of curiosity, why does not r go between 0 and csc(theta)?

RIGHT! Function is undefined for X>0. Therefore, n/2<=theta<=n, and 0<=r<=sin(theta)

That's great! Does this mean that we can write the integral as a sum of two integrals (first quadrant circle and second quadrant circle) 0<=theta<=n/2, 0<=r<=1 + n/2<=theta<n, 0<=r<=sin(theta) ? OR( 0<theta<n (0<=r<=1 + 0<=r<=sin(theta))

Thank you for correction. If (pi/2)<=theta<=pi, does it meant that sin(pi)<=r<=sin(pi/2); 0<=r<=1. that's weird

In the second quadrant = > (pi/2)<=theta<=pi 0<=r<=0.5

So, to get to the right conclusion do I add the area of the half of the circle centered at (0,0.5)?

Explanation: I started out with the graph. In the first quadrant we have a quarter of a circle x^2+y^2=1 (radius = 1. In the second quadrant we have x^2+(y-0.5)^2=0.25 (radius=0.5). Then, since the function exists both in first and second quadrant, 0<=theta<=pi. Then, by my false reasoning, I...

Well, ∫∫(r^2)drdtheta For clarification, http://www.math.utsc.utoronto.ca/b41/oldexams/2004ffinal.pdf q8c All what is required of me is to change the limits of integration what I got is: 0<=theta<=pi 0.5<=r<=1

Homework Statement ∫∫√(x^2+y^2)dxdy with 0<=y<=1 and -SQRT(y-y^2)<=x<=0 Homework Equations x=rcos(theta) y=rsin(theta) The Attempt at a Solution 0.5<=r=1, we get r=0.5 from -SQRT(y-y^2)<=x by completing the square on the LHS then, 0<=theta<=pi But, when I calculated the...