Calculus - Optimization Problem?
925743
This is pretty straightforward.\nFirst you have to write the formula fo rthe cost of the tank as a function of, say, its radius, r:\n\nThe volume of the cylinder is 100 cubic feet.\nIt is also pi*r^2 times the height, h\n\nso let's solve for h\n100=pi*r^2*h\nh = 100/(pi*r^2)\n\nnext, the cost:\nthe rectangle cost is 2.40*(h*2*pi*r),\nor 2.4*100/(pi*r^2)*2*pi*r\nor, simplifying, 480/r\nThe cost of the base and top are:\n1.50*2 pieces * pi*r^2\n\nSo the total cost is\n480/r+3*pi*r^2\n\nNow you need to differentiate this and find the roots, and make sure they are minima, not maxima!
1121843
the given function is not differentiable at x=0\n\ntherefore,   f(x) = x   for x>0        dy/dx = 1\n                           f(x) = -x  for x<0        dy/dx = -1
764897
use derivative rule for exponential as\n\nd/dx(A exp(bx))=(A*b)*exp(bx)\n\nyou got hint. No more Spoon Feeding
711217
Assume the rectangle has dimensions a x b\nDraw a radius to one of the verticies of the rectangle.\nThen drop a perpendicular from the center of the circle to the midpoint of one of the sides of the rectangle.\nThis forms a right triangle with sides a/2, b/2 and hypotenuse 5.\n(see picture: http://mudandmuck.com/str2/optimization.JPG )\n\nSo (a/2)^2 + (b/2)^2 = 5^2 = 25\nor a^2/4 + b^2/4 = 25\na^2 + b^2 = 100\nand b = √(100 - a^2).\n\nNow we want to maximize the area of the rectangle..\nThe area y = ab = a√(100 - a^2)\n\nTake the derivative with respect to a and set = 0.\n\ndy/da =  √(100 - a^2) +  a * (1/2) * (100 - a^2)^(-1/2) * (-2a)\n\ndy/da = √(100 - a^2) - a^2/√(100 - a^2)  = 0\n\nmultiply both sides by √(100 - a^2) to get:\n\n100 - a^2 - a^2 = 0\n2a^2 = 100\na^2 = 50\na = √50 = 5√2\nb = √(100 - a^2) = √50 = 5√2\n\nSo the rectangle is indeed a square of side length  5√2
607760
Well, when you maximize or minimize with constraint, you use Lagrange multipliers. So set that up. Now write the Euler-Lagrange equations for your system. Be sure to consider the boundary carefully!
783536
let the length and width be 'x' and 'y' reply.\ngiven xy=512\nso x=512/y\nthe perimeter=2x+y\n=2x+512/x\nfor optimising\ndP/dx=0\n2-512/x^2=0\nmultiplying by x^2\n2x^2-512=0\n2x^2=512\ndividing by 2\nx^2=256\nx=rt256=16\ny=512/x=512/16=32\nso the dimensions for min perimeter\n=16'*32'\nperimeter=2x+y=64'\ncost=64*2=$128
867789
Try starting with a proportion. Let K be the cost we are looking for.\n\nb / x = K / y\n\nNow solve for K.\n\nK = by / x
861163
First you graph the feasible region.  Let x be the number of burritos sold, y the number of nachos.  10<= x <= 40.  This gives you a vertical band, shading between a vertical line crossing the x-axis at 10 and one crossing the x-axis at 40.  Likewise 30<= y <= 70 gives you a shaded band between horizontal lines that cross the y-axis at 30 and 70.  Where they overlap you have a rectangle.  The final constraint is x+y <= 90, which slices off the top right corner of the rectangle.\n\nNow you need to know the coordinates of the corners.  Bottom left is (10,30), top left is (10,70), bottom right is (40,30), upper top right is (20,70) and lower top right is (40,50).\n\nFinally, your profit function is P = 0.95x + 0.45y.  It's slope is a little steeper than -2, and will reach a maximum at a corner of the feasible region far from (0,0).  \n\nTry it at the upper top right: .95(20)+.45(70) = 50.50\nTry it at the lower top right: .95(40)+.45(50) = 60.50.  So max profit with 40 burritos and 50 nachos.
902349
Hynton has it.  Just remember that you'll be restricted to x > 0 because of the original form of the problem.
435388
damnt i should have paid attention calculus...\n\nlet pam...pam...breat....dimension......carry the one.....as x approaches zero......carry  the two....hold on...\n\nthis could take a while.
932642
Initial concentration is C(0) = 3[e^0 - e^0] = 0\n\nFind maximum.  1) Differentiate, 2) set to zero, 3) find t at this point.\n\n1) dC/dt = 3[-0.5*e^(-0.5t) + 0.6*e^(-0.6t)]\n\n2) 3[-0.5*e^(-0.5t) + 0.6*e^(-0.6t)] = 0\n\n3) 0.6*e^(-0.6t) = 0.5*e^(-0.5t)\n\n0.6 = 0.5*e^(-0.5t)/e^(-0.6t)\n\n0.6 = 0.5*e^(0.1t)\n\n1.2 = e^(0.1t)\n\nt = ln(1.2)/0.1 = 1.8232155679395462621171802515451
765864
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
347106
Okay, if you're actually serious, you REALLY should be paying more attention in class, but I'm gonna answer it for two points anyway.\n\nThe derivative of x^2 (with respect to x) is 2x.\n\nHave fun.
612056
the local maxima can be found at the points where dy/dx is equal to zero and (d^2.y/(dt)^2) is less than zero. The local minima can be found at the points where dy/dx is equal to zero and (d^2.y/(dt)^2)(read as dee sqared y by dee tee squared) is greater than zero.\n\nThe explanation is that if u were to draw the graph of watever function u want to find the maxima and minima of,.... at the point where its max, the tangent is zero and the differential of the tangent is sloping downward!! And it slopes upward for the point of minima....\n\nHope this helps.\n\nYou better gimme best answer for this.
437876
Take the derivative:\n\nf'(x) = 2x - 4\n\nSet the derivative equal to 0, and solve for x:\n\n0 = 2x - 4\nx = 2\n\nSub this value of x into the original equation to find the turning point:\n\nf(2) = (2)^2 - 4(2) + 1\nf(2) = -3\n\nSo, your turning point is (2, -3).\n\nAnd, since the x-squared term in the equation is positive, you know that this point is a minimum.\n\nDone and done.
846527
You'll have to do this in more than one step.  Basically it will sort of have to be a program.  I would start with the restart; command and then  define the function (I would look this up in the help menu since it is dependent on 2 variables).  Then use a command to find the derivative.  Then a command to solve for the pairs (x,y) when the derivative is 0.  That will give you 1.  For 2 and 3 you will also have to do a series of commands.  Use your calc book to refresh your memory on the procedure of finding these, it should be a similar procedure in maple.
428665
This one is fairly clear.  Hope it helps.\n\nhttp://www.math.hmc.edu/calculus/tutorials/substitution/