Calculus Optimazation?
915466
Let a and b the lengths of the sides of the rectangle (with a the side with the semi-circle).\nThe perimeter is 2b+a(pi/2+1)=16, and the area ab+pi/8*a^2. Substituting b in the area (from the perimeter equation), the area is 8a-a^2*(1/2+pi/8). \nSetting the derivative to 0, we have a=32/(4+pi). Then b comes from the perimeter constraint as b=8*(1-(2+pi)/(4+pi)).\nNumerically this gives a=4.48, b=2.24 and an area of 17.92
405573
you need to use some calculus.\nget an expression for the area \ndifferentiate it\nthen find when it is a maximum.
896473
Here's the method (do the calculations yourself).\n\nP(x) the price at which you can sell the quantity x\nR(x) = xP(x)  [we don't know the price yet, it depends on the unknown x]\nProfit = R(x) - C(x) = xP(x) - C(x)\n\nTo maximize the profit over x, work out its derivative with respect to x, equate it (the der.) to 0, and solve for x. That's the optimum (profit-maximizing) quantity.\n\nIf you also want to know the price and the maximum profit, put the x you've found into the formula for P(x) = ... . Finally, put this x into the formula for the profit: it is xP(x) - C(x) = ...
881328
this requires way more than anyone can do here, but, i know of a cool java program that will do it for you: \n\nhttp://www.math.rutgers.edu/courses/ODE/sherod/phase-local.html\n\nyou can even set your initial conditions and print it out!
1029299
I just took Calculus this year.  And I apologize in advance, since Calculus is not my strong suit and it is especially difficult to explain without visuals.  So....\n\nTo find the relative min and max, you take the first derivative of the function.  You have to multiply the exponent by the coefficient and then subtract one from the exponent.  Therefore, the first derivative of the function is:\nf'(x) = -4x^2+2x+30\n\nThen, you set this equation equal to zero to find the relative mins and max.  What I like to do is use a graphing calculator to graph the equation of the first derivative.  See where the line crosses the x axis.  The points where the graph go from positive to negative are your rel. max, and the points where the graph goes from negative to positive are your rel. min.  For this one, your relative min is at x=-2.5, and your rel. max is at x=3.\n\nI'm not sure about this part, but I believe that the graph of the first derivative is also good for telling you how to find out where the graph is increasing and decreasing (I think).  Now, the graph of the first derivative is positive in between -2.5<x<3.   Which means that is where the original function is increasing.  \n\nNow on to the inflection points.  You find them by taking the second derivative.  Take the derivative of the derivative (see directions above).  The second derivative is\nf''(x) = -8x+2\n\nThe inflection points are where x=0 on the graph of the second derivative.  I think you're set.  \n\nAnyways, if this didn't help, a real teacher will do wonders with this problem and many others, I know mine did! (If you're reading this, thanks Mrs. C!)  Good Luck!
657050
1) f'(x) = 40x^4+6x^2+4\nTo find local max's and min's, set this equal to zero, but you will see that all solutions will be imaginary, so there are no local max's and min's on the real line.\n\nf''(x) = 160x^3+12x\nTo find the points of inflect, set this equal to zer, but you will see that the only place where there is an inflection is a x = 0.\n\nSo what have we learned?  f(x) is a strictly increasing function, meaning tha there could only be one real root.\n\n2. Let u = 1+t and use substituion.
1061316
Calculus tells us a lot about optimizing functions. However, it isn't needed here. No matter what planet you're on, the height-vs-time graph of a free-falling object is always a parabola.\n\nThere are a lot of facinating facts about parabolas and conic sections in general, but let's get to answering the questions that you asked. ;)\n\nThe maximum or minimum of a parabola occurs at its vertex. The value of the independent variable ( t ) at the vertex of a parabola is -B/2A. Therefore, t = -16/2(-4) = -16/-8 = 2.\n\nA) The bullet reaches its highest point two seconds after it is fired.\n\nMoving on, the height of the bullet it determined by the formula itself: h = -4t^2 + 16t + 5. If the bullet is on the ground, then h = 0. So:\n\n-4t^2 + 16t + 5 = 0\n4t^2 - 16t - 5 = 0 [Multiply by -1 to work with leading positive.]\nt = [16 +/- sqrt(16^2 - 4(4)(-5))]/(2*4) [Quadratic Formula]\nt = [16 +/- sqrt(336)]/8\n\nThe negative root does no good, as the bullet cannot land before it was fired. But the positive root show us that:\n\nB) The bullet hits the ground approximately 4.3 seconds after is was fired.\n\nLast is actually the easiest part, although most students miss it. You know the bullet reaches its highest point when t = 2 from part A. What will be its height at that point? You already have the equation. h = -4t^2 + 16t + 5.\n\nh = -4(2)^2 + 16(2) + 5 = 21\n\nC) The bullet will reach a height of 21 feet.\n\nSo, to sum up:\n\nA) The bullet reaches its highest point two seconds after it is fired.\nB) The bullet hits the ground approximately 4.3 seconds after is was fired.\nC) The bullet will reach a height of 21 feet.\n\nI hope this helps! =]
152408
Use excel to give delta y/delta x on the curve. Look for a point where this value (the derivative) shoots up in value. That's your point.
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x(abs[x- (x^2)] (-1=2)\n\n:> peace\n.
1026879
f(x,y)= (1-x)^2 + 105 (y- x^2)^2\n\nIt's often used to test out various nonlinear optimization algorithms.
794869
Here's a good website that discuss implicit differentiation with some example problems.\n\nHope this helps
888343
Use the chain rule:  dy/dx = d/dx[(ln x)^x] = x(ln x)^(x-1)d/dx[ln x]
57176
http://www.calculus.org/
906303
Brute force though it may be, an x-dimensional linear regression is still your best bet for optimization.  If there is a single optimal process, the method will yield it, but you are more likely ro arrive at a number of processes, each optimized, that will yield a maximum profit.\n\nYour other alternative is to optimize each process independently, and then choose the most profitable, but you are likely to discard combinations which may yield a better end result, and you will still need  n-dimensional regressions of some sort.\n\nIn lieu of dedicated software for the problem, a spreadsheet can be used quite effectively.  For x > 7, however, you may need to apply techniques to keep numbers within your computer's ability, although this is not as problematic as it is with polynomial regressions.
79903
Here is an excellent site:\nhttp://www.sosmath.com/calculus/calculus.html\n\nVery thorough explanations and examples provided.
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Profit- P(x)\n\nP(x) = R(x) - C(x)\nP(X)= 20x - .001x^2 - 2500 - 10x\nP(x)= 10x - .001x^2 - 2500\n\n\nWe have our equation, let's differentiate it (i'll use the shortcut and the power rule to speed things up).\n\nP'(x) = 10 - .002x\n\nThe zeroes of this will be local extrema of the original equation.  Since there is a negative sign in front of the .001x^2, that flips the parabola upside down and it will be the maximum point.\n\n0 = 10-.002x\nx=5000\n\nPlug this into the cost equation:  2500 + 10(5000) = 2500 + 50000 = 52500.\n\nHope this helps.
900462
It really depends on how the equation of the path/curve is originally specified.\n\nIf it's specified in terms of x and y, an easy parameterization is:\n\nx = t\ny = f(x) = f(t)\n\nBut there are many other ways to do it.