optimization problem, maximizing perimeter?
826437
(x/a)²+(y/b)² = 1 (← Note my variation! I am using a², b² in place of your a, b)\n\nx = acos t , y = bsint is its parametric equation, a, b >0\n\nThen if T1(acost, bsint), (0 ≤ t ≤ π/2),  is the point on the ellipse in the 1st quadrant, the point on the ellipse vertically below it is\nT2 (acost, -bsint).\nThen the point horizontally to the left of that on the ellipse is\nT3 (-acost, - bsint)\nThe final point is vertically above on the ellipse namely\nT4 (acost, -bsint) which is also horizontally to the left of the point T1\n\nSo perimeter P = |T1T2| + |T2T3| + |T3T4| + |T4T1|\n= 2bsint + 2acost + 2bsint + 2acost\n= 4(bsint + acost)\n\ndP/dt = 4(bcost - asint)\n= 0 for stationary points\nSo asint = bcost\nie sint/cost = b/a\nSo t = arctan(b/a) (0 ≤ t ≤ π/2) (ie a,b>0)\n\nd²P/dt² = -4(acost + bsint)\n< 0 for (0 ≤ t ≤ π/2)\nand therefore P is maximum when t = arctan(b/a) (0 ≤ t ≤ π/2)\n\nIf t = arctan(b/a) then tan t = b/a \nThus cost = a/[√(a² + b²)] and sint = b/[√(a² + b²)]\n\nThus T1 ≡(a²/[√(a² + b²)], b²/[√(a² + b²)]), \nT2 ≡(a²/[√(a² + b²)], -b²/[√(a² + b²)]),\nT3 ≡(-a²/[√(a² + b²)], -b²/[√(a² + b²)]),\nT4 ≡(-a²/[√(a² + b²)], b²/[√(a² + b²)])
807703
Achieved Theoritical Upper Limit
360732
Right.  I'm not sure what the question is, but let's assume that the space we are talking about is a rectangle, and you have to figure out what the dimensions are.  Well, the perimeter, 26, is the distance around the space which has the area of 36, which for the sake of argument, we will assume is square feet.  What times what equals 36?  9 x 4, right?  So, it's possible, but not certain that we are looking at a rectangle whose dimensions are 9 feet by 4 feet.  To prove this, remember that a rectangle has four sides.  If our assumptions are correct, it has two sides of 9 feet and two of 4 feet, right? So then,  9 + 9 + 4 + 4 = 26.  It would seem that this is the right answer.  \n\nOf course 12 times 3 also equals 36.  But this time, if we substitute the 12 and the 3 as the dimensions, we would get a perimeter of 30 feet, obviously the wrong answer.  Looks like you have a rectangular space that is 9 by 4 feet.\n\nI hope this helps and that I haven't just muddied the water further.
1015708
Do you mean a "fixed perimeter?"
343312
perimeter = 2 (side1 + side2)\narea = side1 * side2
800206
The 400 feet of fencing forms the perimeter P of the rectangular area.  The perimeter of a rectangle is P = 2*W + 2*L.  The area of a rectangle is A = L*W.  Take either the W or L from the perimeter equation and solve for it in terms of the other.  For example. 2*W = P - 2*L, or W = (P - 2*L)/2.  Put this expression for W into the area formula\n\nA = L*(P - 2*L)/2\n\nA = P*L/2 - L^2\n\nTo maximize area, take the derivative of A with respect to L to get\n\nA' = P/2 - 2*L  and set it equal to zero\n\nP/2 - 2*L = 0, or  L = P/4;  This gives you the length of the area.  You can get the width from the perimiter formula W = (P-2*L)/2;\nPutting in L = P/4 gives W = (P - P/2)/2 = P*(1-.5)/2 = P/4.\n\nNotice that L = W = P/4; in other words the maximum area rectangle is a square.
617970
I am terribly afraid of the equilibrium problem.. the first one on the free response!!! well, good luck!
905517
I answered this question for jc yesterday!  Search for questions to find the answer.\n\n\nJust integrate p(x) along the rod.
114649
Here's an idea.  Show them something that's pretty much flat and rectangular, such as piece of stiff carpet or a pad of paper.  Write on the board what the perimeter, area, and volume are.  Have them measure it with a tape measure and come up with the answers that you have.  They'll see that they just have to add for the perimeter but multiply for the area and volume.
897940
perimeter = 2 * (length + width)\n\nfor square, length = width..\n\nso, perimeter = 2 * (2 * length) = 4 * length\n\ngiven, 4 * length = 104\nso, length = 26\n\nso, each side is 26 cm long.
634292
It depends on what you want to optimize.\n\nIf you want time to production optimized, assign weights based on the activities ability to delay the timeline.\n\nIf you want cost to be optimized, weight based on time and cost functions.
724798
Hints on the first two:\n\n1. Minimize A = 2 * pi * r^2 + 2 * pi * r * h subject to pi * r ^ 2 * h = 1000.\n\n2. Minimize A = 8 * r^2 + 2 * pi * r * h subject to pi * r ^ 2 * h = 1000.
802897
My main problem is 'entouring'  I presume that you mean surrounding but I can see no reason for it. If you already have a 5m by 5m square enclosure then why waste fencing going around the enclosed sides? If you do you have only 7metres length of fencing left on one of the box sides to bow out to enclose a bit of grass and it would be much more sensible to terminate the ends of your fence at the ends of one face of the box leaving an area for grazing which is that part of a circle with an arc of 22m and a chord of 5m length formed by the box.  Leaving aside the practicalities of the problem ( need for a lot of fence posts to get a partly circular fence - the fact that sheep are of finite thickness so can't squeeze through small gaps or graze in very thin areas ) then a circle of radius roughly 4.35metres almost meeting at the corners of one face of the box will do the best job.\n\nAh! - added later : \nso you need to solve simultaneous equations, something like  r*theta=7 and r*sin(theta/2)=5/2 \n\nThere's something in Euclid about the length of a chord which might be more direct but I can't remember it. \n\ntheta = sin(theta/2)*14/5  and you should be able to expand sin(theta/2) in a series involving theta\nsin(x) = x −x^3/3! + x^5/5! - ...... and solve aproximately or use some identity involving e^(ax) \n\nsin(x) = 0.5i*[ e^(ix) - e^(-ix) ]\n\nwhich leaves you solving for 10/14*theta = ie^itheta/2 - ie^(-itheta/2) . hmm \n- check out euclid or solve graphically to get r~=2.55metres, theta about 1.3725*2 radians.
751869
First, we need to find the partial derivatives of cost with respect to length and width.\n\nC = 10LW + 640(1/W + 1/L)\ndC/dL = 10W - 640 * (1/L^2)\ndC/dW = 10L - 640 * (1/W^2)\n\nThe minimum of cost will occur where the above partial derivatives are equal to 0.\n\n10W - 640 * (1/L^2) = 0\n10W = 640 * (1/L^2)\nW(L^2) = 64\n\n10L - 640 * (1/W^2) = 0\n10L = 640 * (1/W^2)\nL(W^2) = 64\n\nSo we have 2 equations which both need to be true at minimum cost\n\nW(L^2) = 64 and L(W^2) = 64\nL = 64/(WL) and W = 64/(WL)\n\nSo W and L must be the same\n\nW(W^2) = W^3 = 64\nW = L = 4 meters\nH = 2 cubic meters / (4 * 4 square meters) = 0.125 meters\n\nNote that the given equation is correct, Gary needs to substitute\nhw = 2/l and hl = 2/w to get the final result...
87402
a: width\nb: length\nperimeter = 2 x (a+b)= 420 ---> a+b = 210\nwe need to find max (a x b)\n\na= 210-b ---> area = a x b = b x (210-b) = 210 - b2 = - b2 + 210b = - b2 + 2 x 105 x b - 105squared + 105squared\n= - (b2 - 2 x 105 x b + 105squared) + 105 squared\n= - (b - 105)squared + 11025\n\nSince (b - 105) squared is always >= 0,\n - (b - 105)squared is always <= 0\n  - (b - 105)squared + 11025 is always  <= 11025\n\nthe largest area will be 11025 squared feet. This happens when - (b - 105)squared = 0, which means b = 105.\n\na = 210 - b = 105\n\n\nthe shortest way is to say that a rectangle has the largest area when it's a square, so each side would be 420/4 = 105.
1090901
In this problem, you simply substitute "a" and "b" into the equation. The answer is 110 mm. Work shown:\n\n4a + 2b = perimeter\na = 10 mm\nb = 35 mm\n4(10) + 2(35) = perimeter\nPerimeter = 110 mm\n\n:)
364028
Richard Bellman made this principle in regards to a two-stage optimization to get V(t+s, x, φ)\n\n   Using any U up to a time t, with a controller U changes to an optimal one.   At time t+s we get a performance J(t,x,V(s,∙,φ),U).  If you take the supremum with respect to U you get V(t+s, x, φ)