Need Help with Calculus: Optimization?
816542
Find derivatives with respect to each variable.
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
982689
Here's the method (you can do the calculations yourself, surely).\nMaximize the profit xp(x) - C(x) over x: work out d/dx of the profit, set it equal to 0, and work out x. That's the optimum (profit-maximizing) quantity. Then put this x into the formula for p(x) = ... to work out the price. Finally, put this x into the formula for the profit: it is xp(x) - C(x) = ...
890169
x = r( t - sint); y = r(1 - cost)\ndx/dt = r(1 - cost); dy/dt = rsint\n\ndy/dx = (dy/dt)/(dx/dt)\n= (rsint)/(r(1 - cost))\n= sint/(1 - cost)\n\nSo at the point where t = a\n\nSlope of tangent (m) = sina/(1 - cosa)\nWhen t = a. x = r(a - sina) and y = r(1 - cosa)\nSo tangent is of the form\ny = mx + b\nand it passes through (r(a - sina) , r(1 - cosa))\n\nThus requred tangent is of form \n\ny = sina/(1 - cosa) x + b\nAnd r(1 - cosa) = sina/(1 - cosa) * r(a - sina) + b\nb = r(1 - cosa) (1 - cosa)\n= r[(1 - cosa)² - sina(a - sina)]/(1 - cosa)\n= r[ (1 - 2cosa + cos²a - asina + sin²a)/(1 - cosa)\n= r[(2(1 - cosa) - asina)]/(1 - cosa))\n= 2r - arsina/(1 - cosa))\nSo the equation is\ny = sina/(1 - cosa) x + 2r - arsina/(1 - cosa))\nOR\nxsina - y(1 - cosa)  + 2r(1 - cosa) - rsina = 0\n\nb) Tangent is horizontal when dy/dx = 0\nie sint/(1 - cost) = 0\nwhence sint = 0\nSo t = -π, 0, π (for -π ≤ t ≤ π)\n\nBut when t = 0, cos t = 1 \nSo dy/dx is indeterminate there (0/0)\n\nBy d'Hopitals Rule\nlim sint/(1 - cost) =\nt →0\n\nlim cost/(-sint)  → ±∞\nt →0\n\nSo horizontal tangents when t = -π, π\nie at (-πr, 2r) and (πr, 2r)\n\nThe tangent is vertical when dy/dx is undefined,\nie 1 - cost = 0\nSo cost = 1\nie t = 0 (for -π ≤ t ≤ π) and here the slope becomes undefined as t→0 (see above)\n\nSo vertical tangent at t = 0 ie at (0, 0)
826268
2l + 2w = 300\n2l = 300 - 2w\nl = 150 - w\n\nMaximize length times width:\n\nA(w) = w(150 - w) = 150w - w²\n\nTo maximize, take derivative, set = 0, and solve for w.\n\n150 - 2w = 0\n2w = 150\nw = 75\n\nand then l = 150 - 75 = 75.
832985
Let S = x + 1/x.\n\nWe want to minimize x for x>0.\n\nTake the derivative of S with respect to x and set that equal to zero to find the extrema of S:\n\n0 = dS/dx = 1 - 1/x^2\n\n1/x^2 = 1\n\n1 = x^2\nx = 1 (and -1, but we are only interested in x>0 here).\n\nWe can check to make sure this is a minimum by looking at the sign of the second derivative of S with respect to x at x=1:\n\nd2S/dx2 = 2/x^3\n\nWhen x = 1, the second derivative is equal to 2, which is positive, so the extremum at this point is a minimum (the function is concave upward).
589794
Formulate stability of the system as a function of its energy. Now, one has to show that minimum energy translates into max stability on finding the derivatives.
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Problem 1: Find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4). \n\nYou find the y-intercept by replacing x with zero, which gives f(x) [which is y on the graph] = 2, so b + c = 2 (since a^0 = 1)\n\nThe horizontal asymptote is y = c, so c = -2.  That means b must be [you figure it out]\nNow it is y = 4a^x - 2.  This must be true when x=1 and y=4.\nSo 4 = [you figure it out; answer deleted due to complaint from below]\n\n\nProb. 3.  What is the amount in the account if $500 is invested for 10 years at the annual rate of 5% compounded continuously? \n  Just plug in the values.\nA = 500 e^(.05 x 10)\n\n[answer deleted]
409149
If you are looking for only the real values,\n\ndifferentiating once,\nd(x³-6x²+12x-4)/dx = 3x^2-12x+12\n\n3x^2-12x+12 = 0\n\nx^2 - 4x + 4 = 0\n\n(x-2)^2 = 0\n\nx = 2\n\nwill be the critical point;\n\ndifferentiating twice,\n\n2x-4=0\nx=2\n\n+ve constant, so that the function will have a minimum at this point.\n\n\nDaneengineer@yahoo.com
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you need to use some calculus.\nget an expression for the area \ndifferentiate it\nthen find when it is a maximum.
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Find the first derivative and set it equal to zero.\nf'(x) = 6x^2 - 30x + 24\n6x^2 - 30x + 24 = 0\nDivide everything by 6\nx^2 - 5x + 4 = 0\nFactor\n(x-1)(x-4) = 0\nx = 1 or x = 4\nTo test if these are max or min's, we can use the second derivative.\nf''(x) = 12x - 30\nFor x = 1, f''(1) = 12(1) - 30 = -18 < 0\nThis means f(x) is concave down at x = 1; hence, x = 1 is a local max.\nFor x = 4, f''(4) = 12(4) - 30 = 18 > 0\nThis means f(x) is concave up at x = 4; thus, x = 4 is a local min.\n\nIf you are looking for the points, then you will need to substitute x = 1 and x = 4 into f(x) to find the corresponding y value for each.
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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
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Correct, derive and solve for x=0, that will give you the maxima or minima.\nUsing product rule:\ndy/dx= 2ln(x)*(1/x).\nSolve for 2ln(x)/x=0; *x both side\n2ln(x)=0; /2 both sides\nln(x)=0; \nThe answer is 1. The natural log of 1 is zero.\nx=1\n\nTherefore, the min or max is at 1. \n\nIf you want to know if its a max or min you take the second derivative.\n\nd^2y/dx^2= 2/x^2-2ln(x)/x^2=2-0=2 -> If evaluated at the min/max value of x=1.\n\nSince the second derivative is positive, this function is convex, and at 1 it is minimized.
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You want to minimize the length of the line segment running through points (0,A), (8,15), and (B,0).\n\nthis line is represented by the equation:  y = (15/(8-B))x + A\nsolve for A:\nA = y - (15/(8-B))x\nplug in values for x and y using the one known point (8, 15):\nA = 15 - (120/(8-B))\n\nlet f be the length function you want to minimize.\nf(B) = sqrt(B^2 + (15- (120/(8-B)))^2)\n\nnote that you are only interested in values of B>8.\n\nnow you can take the derivative of f using the chain rule and then solve for f'(B) = 0...\nor you can use a graphing calculator and find the minimum on the interval (8, inf).\n\nB should come to 20.164 or so.\nplug this value into f:\nf(20.164) = 32.013, approximately.
869572
Assuming\nX = # of units\nR(x)=Income\nC(x)=Cost\n\nNow profit would equal income - cost\nP(x) = R(x)-C(x)\nP(x) = 2x - (0.01x^2 + 0.6x +30)\n        = -0.01x^2 + 1.4x -30\nUsing calculus or some way to find the maximum of the binomial function P(x) or just graph it and find maximum.\n\nProfit is maximized at x = 70 units\nProfit = -0.01*70^2 +1.4*70 - 30 = $19
888459
v(t)= -gt - ve(ln((m - rt)/m))\n\nx(t) = ∫v(t) dt\n\n= ∫[-gt - ve(ln((m - rt)/m))] dt (Note ∫lnu du = ulnu - u + k)\n\nLet u = (m - rt)/m Thus du = -r/m dt ie dt = -m/r du\n\nSo x(t) = -½gt² - ve*-m/r (u(lnu - 1) + c\n= mve/r [ (m - rt)/m * (ln((m - rt)/m)  - 1] - ½gt²\n= ve(m - rt)/r * (ln((m - rt)/m) - 1) - ½gt²\n\nIf g = 9.8 m/s^2, m = 30000 kg, r = 155 kg/s, and ve = 3000 m/s and t = 1 minute = 60 s\n\nx(60) = 3000 * (30000 - 155 * 60)/155 * (ln (30000 - 155 * 60)/155 - 1) - ½ * 9.8 * 60²\n\n≈ 138390 m\n\n= 138.39km
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Here is an excellent site:\nhttp://www.sosmath.com/calculus/calculus.html\n\nVery thorough explanations and examples provided.