I need help with Calculus more specifically with Optimization?
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2x^3 + x - 2 has only one real root and it is positive.\n(I used Decartes Rule of Signs to figure out the number of possible roots.)\nUsing my graphing calculator I get x = .83512235
845625
y = -x^2 + 6x - 5.5\ndy/dx = -2x + 6\nInsist dy/dx = 0 to find local maximum.\n0 = -2x + 6\n2x = 6\nx = 3\ny = -(3)^2 + 6(3) - 5.5\ny = -9 + 18 - 5.5\ny = 3.5\n\nThe vertex is at (3, -3.5).\n\nIf you have a polynomial equation of the form,\ny = a x^n + b x^(n-1) + ...\nThen the first derivative is\ndy/dx = an x^(n-1) + b(n-1) x^(n-2) + ...\nWhen dy/dx is zero, the graph of y(x) has either a maximum or a minimum.  In general, dy/dx is the slope of the line tangent to the curve of y(x) at x.\n\nThe neat thing about using differential calculus to find maxima and minima is it works for just about all (differentiable) functions, including polynomials and trigonometric functions.  Once you know this trick, you can nearly always find those local maxes and mins. Duke95's answer is a special case that can be derived from the more general differential calculus method.
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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)
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Mathworld.com.  Or, if you type the subject your interested in followed by "Forum" in google, you'll find some good help that way.  Fore xample, "calculus forum".  Also, you can ask me whatever you want.
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use logarthmic differentiation\n\ny = x^(1/x)\n\nln(y) = (1/x)* ln(x)\n\ndiff w.r.t x\n\n(1/y) (dy/dx) = (1/x)*(1/x) +ln(x)*(-1/x^2)\n\n(dy/dx) = { y/(x^2) }  [ 1 - ln(x) ]\n\nsubstitute for y
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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! =]
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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) = ...
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well first you have to get the first derivative\n\nso: the 1st derivative= t*cost+1*sint\n                                  = cost^2+sint\nthen you take the derivative of that\nso: the 2nd derivative= 2cost*-sint+cost\n                                   = -2sintcost+cost
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You would need to know how to integrate (that is, find anti-derivatives). The proceedure forthis type of problem can usually be understood after first semester calculus. The specific integrations may need some second semester calculus.
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well optimization is used by companies to get the most amount for their money. (highest volume using lowest surface area). theres differential equations that model population and waves and transients in circuits. \nif you wantto be technical, then quantum mechanics describes everything using a differnetial equation (schrodinger wave eq).\num 1 of the easiest would be measuring speed and acceleration. as in if you know your position over time, then you can find your speed by differentiating and acceleration by differentiating again.\n\nlook in any calculus text book at the questions and there are a lot that give you a real world situation. i dunno. hope that helped.
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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!
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You need to differentiate the equation y=2x^2. \n\ndy/dx = 4x therfore the gradient at x=3 = 4*x = 12
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I just taught this section today to my calc students.  You start by finding the first and second derivative.  I recommend looking in your book for an example.  Your book probably has something like this towards the front of the section.  It would be too long winded to write everything down here.  If you don't have your book on you, the web page below has problems that you can do and then check your answer.\n\nhttp://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html
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Ha ha! I used that book myself. Check this website out, it may be what you're looking for.\n\nhttp://math.cramster.com/calculus-5th-solutions-10-202.aspx\n\nSolutions manuals are great if you get stuck and just need a little hint, but don't depend on them. You'd probably be better off just finding someone who knows how to do the problem and can help you with it than looking at the solutions manual all the time.\n\nBut hey you're an adult, I think, and can make your own decisions.
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This one is fairly clear.  Hope it helps.\n\nhttp://www.math.hmc.edu/calculus/tutorials/substitution/
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Try "Calculus" - By Gilbert Strang\nThis is available along with Student study guide &\nInstructor's manual!\n\nhttp://ocw.mit.edu/ans7870/textbooks/Strang/strangtext.htm
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To find the direction of maximum ascent, find the directional derivative (DD), and find the direction where it has its max.\nNote: This is the equation of a plane, so the directional derivative is independent of the point where it is taken.\n\nIf v is a unit vector, then the DD in the direction of v, is the vector dot product of v with the vector  formed from the two partial derivatives\nPartial Z with respect to x is 1.5\nPartial Z with respedt to y is -.8\n\nDD = (1.5, -.8) dot v\n\nSince v is  unit vector, you can represent it\n\n(sin t, cos t)\n\ngiving us DD = (1.5)sint - (.8)cost \n\nTo find direction the max, take the derivative with respect to t, and set to 0.\n\n0 = (1.5)cost + (.8)sint\n-(1.5)cost = .8sint\n-1.5/.8 = tan t\n\nThe solutions to that say that v has the same direction as either of the (non-unit) vectors\n(1.5, -.8) or (-1.5, .08). The first is maximum ascent, the second maximum descent.