8 x 2 1 2 x 2 y 2 0 x y 2 1 2 y 2 Explanation Since we have no variables outside the parenthesis, we can just multiply the 4 by every More Items ShareThe trace in the x = 1 2 plane is the hyperbola y2 9 z2 4 = 1, shown below For problems 1415, sketch the indicated region 14 The region bounded below by z = p x 2 y and bounded above by z = 2 x2 y2 15 The region bounded below by 2z = x2 y2 and bounded above by z = y 7A point where f00(a) = 0 and f000(a) 6= 0 is called a point of in°ection Geometrically, the equation y = f(x) represents a curve in the two
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F x y z x 2 y 2 z 2 x 4 y 4 z 4 1-Equating these pair wise and simplifying we nd x 2= y, x2 = z2 and y 2= z implying that x= yand y= z Plugging this into g(x;y;z) gives 3z4 = 1 so that z= 3 14 Combining this we nd that the collection of points is then (4) P 3 = f( z;Apr 23, 12 · Use the Lagrange Multipliers to find the maximum and the minimum values of the function f(x,y,z)=x^4y^4z^4?
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Nov 21, 08 · then x^2 y^2 = 18 we need z^2 to be less than this, but still as big as possible so let's let z = 4 (so that z^2 = 16, which is pretty close)** with these numbers, x^4 y^4 = 162, which is much less than z^4 = 256 therefore, NO to the prompt question, so, insufficientF(x,y,z)=x^2y^2z^2 subject to g(x,y,z)=x^4y^4z^4=1 Fx = 2x Fy = 2y Fz = 2z Gx = 4x^3 Gy = 4y^3 Gz = 4z^3 Fx = lamb * Gx 2x = lamb * 4x^3 1 = lamb * 2x^2 view the full answer Previous question Next questionExample 1 Find the maximum and minimum values of the function f(x;y;z) = x2y 2z subject to the constraint x4y4z4 = 1 (Exercise #11 in Stewart, x158) Solution 1 Let g(x;y;z) = x4 y4 z4 Then the constraint is g(x;y;z) = 1 Note that the level set g(x;y;z) =
X 4 y 4 z 4 = 1 BuyF(x;y)dA= Z x=4 x=0 Z y=y max(x) y=0 xdydx (Fubini) = Z x=3 x=0 Z y=x y=0 xdydx Z x=4 x=3 Z y=4x x2 y=0 xdydx = Z x=3 x=0 x2 dx Z x=4 x=3 Z x(4x x2)dx = 9 4 3 x3 1 4 x4 x=4 x=3 = 175 12 If we regard this region as horizontally simple instead, so the xintegral is inside, then the left boundary is always given by x= yand the right boundaryAnd we get x = 0 or x2 = 1 2λ, y = 0 or y 2 = 1 2λ, and z = 0 or z 2 = 1 2λ The point (0,0,0) does not satisfy the side condition If two variables are zero and one is not, say x 6= 0 ,y = z
(2 2 x 2 • y 2) (x 2 y 2 z 2) 2 Step 2 21 Evaluate (x 2 y 2z 2) 2 = x 4 2x 2 y 22x 2 z 2 y 42y 2 z 2 z 4 Trying to factor by pulling out 22 Factoring x 4 2x 2 y 2 2x 2 z 2y 4 2y 2 z 2z 4 Thoughtfully split the expression at hand into groups, each group having two terms Group 1 2y 2 z 2y 4 Group 2 2x 2 y 2 2xA unit vector normal to the surface is n = 8 x i 2 y j 2 z k 64 x 2 4 y 2 4 z 2 = 4 x i y j z k 16 x 2 y 2 z 2 From z x = − 4 x 4 − 4 x 2 − y 2, z y = − y 4 − 4 x 2 − y 2 we obtain dS = 2 1 3 x 2 4 − 4 x 2 − y 2 dAIn this form, we can plug in (1), (2), and (3) into (4) This gives us 1 2 2 1 2 2 1 2 2 =1 From this, we can solve for to get
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Question Use Lagrange Multipliers To Find Max/min Of F(x,y,z) = X^2y^2z^2 With The Constraint X^4 Y^4 Z^4 = 1 This problem has been solved!† x = a is a minimum if f0(a) = 0 and f00(a) > 0;424 Verify the continuity of a function of two variables at a point;
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X 4 y 4 z 4 = 1 My solution As we do in Lagrange multipliers I have considered ∇ f = λ ∇ g where g ( x, y, z) = x 4 y 4 z 4 and the last equation is equivalent to the system of equations { 2 x = 4 λ x 3 2 y = 4 λ y 3 2 z = 4 λ z 314) Objective function \(f(x, y, z) = x^2 y^2 z^2\) Constraint \(x^4y^4z^4=1\) 15) Objective function \(f(x, y, z) = x^2 y^2 z^2\) Constraint \(xyz=4\) In exercises 1621, use the method of Lagrange multipliers to find the requested extremum of the given function subject to the given constraint 16) Maximize \(f(x,y) = \sqrt{6 x(xyz)²=x²y²z²2xy2yz2zx So x²y²z² = 2(xyyzzx) So (x²y²z²)²=4(x²y²y²z²z²x²2xyz(xyz)) {xyz=0) (x²y²z²)²=x⁴y⁴z⁴
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Hi, Different ways to solve this system EQs lend to elimination, substitution methods xyz=4 xyz=6 x yz=4 x yz =6 multiplying 2nd Eq thru by 1 to eliminate both x and y2 z = ∇g(x;y;z) Notice that the second component gives 2y = 2 y So it is natural to break into cases based on whether = 1 or notStack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Visit Stack Exchange
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01 Reminder For a function of one variable, f(x), we flnd the local maxima/minima by difierenti ation Maxima/minima occur when f0(x) = 0 † x = a is a maximum if f0(a) = 0 and f00(a) < 0;FdS, for F(x;y;z) = xyiyzjzxk, where Sis the part of the paraboloid z= 4 x2 y2 above the square 0 x 1, 0 y 1, with upward orientation z= g(x;y) = 4 x 2 y 2 ,物理数学基礎演習演習問題 （平成29年度版） 1 偏微分 11 偏微分係数と偏導関数 問題11 次の2 変数関数z = f(x;y) の1 階偏導関数を求めよ． (1) z = x3 4x2y xy 3y2 (2) z = y
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The density function is λ(x,y) = 13x y 4 Express Z Z Z V f(x,y,z)dxdydz, where V is the solid bounded by the surfaces z = 0,z = y,x2 = 1 − y, as an iterated integral in six different ways 5 Use spherical coordinates to find the volume of the solid that lies above the cone z = p x2 y2 and below the sphere x2 y2 z2 = z 6 Use(2)To come up with this parameterization, rewrite x2 4 y2 = 4 as x 2 2 y2 = 1 and then use x 2 = cos t, = sin It's easy to check that it's reasonable if we plug in x = 2cost, y = sint, and z = 3, then the equations x2 4y2 = 4 and z = 3 are indeed satis ed 2423 State the conditions for continuity of a function of two variables;
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Z 4 1 Zp x x 2 Z x 0 (z xy) dzdydx (c) Z 2 0 Z y2 y2 Z x 0 (z xy) dzdxdy (d) Z 4 0 Zp x x 2 Z x 0 (z xy) dzdydx (e) Z y2 y Zp x x 2 Z xy 0 xdzdydx SolutionSolution The region Dis enclosed by the curves x= y2, x= y2, y= 0, and y= 2 The curve x= y2 is to the left and x= y2 is to the right Since the plane z= xlies above the xyplane forSubject to the given constraint g(x,y,z) = x^2 y^2 z^2 1 = 0 Also find the points at which these extreme values occurLearning Objectives 421 Calculate the limit of a function of two variables;
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Minimize the function f (x,y,z) = x^2 y^2 z^2 subject to the following constraint x y z 24 = 0 Assume Answered by a verified Math Tutor or Teacher We use cookies to give you the best possible experience on our websiteWhere D is the solid region bounded below by xyplane and above by the hemisphere z = p 9 x2 y2 Problem 4 Evaluate ZZZ D xdV;2 (6 points) The boundary of a thin plate consists of the semicircles y = √ 1−x2 and y = √ 4−x2 together with the portions of xaxis that join them Find the center of mass of the thin plate if the density function is ρ(x,y) = 5
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F ( x , y , z ) = x 2 y 2 z 2 ;425 Calculate the limit of aCompute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music WolframAlpha brings expertlevel knowledge and
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Where D is the solidThe points (x,y,z) of the sphere x 2 y 2 z 2 = 1, satisfying the condition x = 05, are a circle y 2 z 2 = 075 of radius on the plane x = 05 The inequality y ≤ 075 holds on an arc The length of the arc is 5/6 of the length of the circle, which is why the conditional probability is equal to 5/6Jul 09, 17 · x^4y^4z^4 = 25/6 Given { (xyz=1), (x^2y^2z^2=2), (x^3y^3z^3=3) } The elementary symmetric polynomials in x, y and z are xyz, xyyzzx and xyz Once we find these, we can construct any symmetric polynomial in x, y and z We are given xyz, so we just need to derive the other two Note that 2(xyyzzx) = (xyz)^2(x^2y^2z^2) = 1 So xyyzzx = 1/2 Note that 6xyz = (xyz
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Jan 31, 09 · stat1 x^2 y^2 = z^2 squaring & solving we get x^4 y^4 > z^4 2*x^2*y^2 now there are 2 possibities 1) x^4 y^4 > z^4 & this condition satisfies the above equation 2) z^4 2*x^2*y^2 < x^4 y^4 < z^4 & this condition satisfies the above equation similarly we can derive the same logic from stat2 hence IMO E Thanks!Z)jz= 3 1 4 g To determine the maximum and minimum values of f(x;y;z) subject to thisIn general f(x,y,z) = x2y C(y,z) will have the desired property for any function C(y,z) Hence, the most general form of f with the property that ∂f ∂x = 2xy is f(x,y,z) = x2y C(y,z) and it is therefore the indefinite integral We now return to the problem of finding the potential
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You can put this solution on YOUR website!=1 f= y,xz,y g= y, x,0 If one of the variables is and the other two are not, then the squares of the two nonzero coordinates are equal with common value and the corresponding value isX y z = 1 Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint (If a value does not exist, enter NONE) f(x,y,z) = x2 y2 z2;
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F x = g x f y = g y g(x;y) = 1 as (1) 2x = y (2) 2y = x (3) xy = 1 Because xy= 1, we know neither xnor yis zero It then follows from either (1) or (2) that 6= 0 (If = 0, then x= 1 2 y= 0, which contradicts to what we know in the previous sentence) Therefore, it is legitimate to divide (1) by (2) to get 2x 2y = y x;X 4 y 4 z 4 = 1 more_vert Each of these extreme value problems has a solution with both a maximum value and a minimum valueUse Lagrange multipliers to find the extreme values of the function subject to the given constraint f ( x , y , z ) = x 2 y 2 z 2 ;
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4 This function is continuous provided it is defined Its domain of definition must have y > 0 and and x2 z2 > y2, so the f(x,y,z) will be continuous on D = {(x,y,z)y > 0,x2 y2 > z2} which is the exterior of a cone with base at the origin, centeredNamely x 2= y From thisFeb 27, 13 · (ii) If x = y = 0, then z = ±1 by using g (Similarly for x = z = 0, etc) (iii) If z = 0 (and x, y nonzero), then we need to optimize f = x^4 y^4, subject to g = x^2 y^2 = 1 By Lagrange Multipliers (again!), ∇f = λ∇g yields = λ ==> x(2x^2 λ) = 0, y(2y^2 λ) = 0 ==> x^2 = y^2 = λ/2 (since x, y are nonzero)
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Math 5 HWK 19b Solns continued §162 p751 This gives us Z R xydA = Z 1 0 1 2 x(1− x)2 dx = 1 2 x(1− x)2 − 1 24 (1−x)4 1 0 = 1 24 the same result as obtained above If you choose to shoot horizontal arrows, integrating first with respect to x and then with respectX 4 y 4 z 4 = 1 Each of these extreme value problems has a solution with both a maximum value and a minimum value Use Lagrange multipliers to find the extreme values of the function subject to the given constraint 11 f ( x , y , z ) = x 2 y 2 z 2 ;422 Learn how a function of two variables can approach different values at a boundary point, depending on the path of approach;
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MATH243 Final Exam Fall 03 2 2 (a) Consider f(x,y,z) = x2y y3z xz3 and the point P = (2,1,−1) Find the rate of change of f at P in the direction of fastest decrease of f at P (b) Given z = y f(x2 − y2), and that f is a differentiable function of one variable, show that yI collected a solution Need to prove x 2 y 2 z 2 5 x y z y = 3 x 4 Arithmetic 699 * 533 6 9 9 Simultaneous equation \left \begin{cases} { 8x2y = 46 } \\ { 7x3y = 47 } \end{cases} \right {8 x 2 y = 4 6 7 x 3 y = 4 7F(x,y)dxdy In our case V = Z 2 0 Z 1 0 (4−x−y)dxdy = Z 2 0 h 4x− 1 2 x2 −yx i 1 x=0 dy = Z 2 0 (4− 1 2 −y)dy = " 7y 2 − y2 2 # 2 y=0 = (7−2)−(0) = 5 The double integrals in the above examples are the easiest types to evaluate because they are examples in which all four limits of integration are constants This happens when the
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Answer to Use Lagrange multipliers to find the maximum and minimum values of f ( x , y ) = x 2 y , subject to the constraint x 2 y 2 = 3 ByX4 y4 z4 = 1 (maximum) (minimum) Assignment DetailsUse Lagrange multipliers to find the maximum and minimum values of the function subject to the given condition f ( x, y, z) = x 2 y 2 z 2;
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Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, historyZ 1 y 3 16 y 2 3 2 y 2 3 2 dy Z 1 1 3 y 16 y 2 3 2 y 4 dy 1 3 1 5 16 y 2 5 2 y from CALC 2302 at University of Texas, Dallas This preview shows page 56 59 out of pagesZ 4 0 Z 2 x2=8 0 Z 1 x=4 0 f(x;y;z)dydzdx Problem 2 Sketch the solid whose volume is given by the iterated integral Z 1 0 Z 4(1 x) 0 Z 2 y2=8 0 dzdydx Problem 3 Evaluate the integral ZZZ D xyzdV;
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Solution We can just minimize the squared distance f(x;y;z) = (x 4)2 y2 z2 subject to the constraint g(x;y;z) = x2 y2 z2 = 1 and then take the square root We have ∇f(x;y;z) = 2(x 4);2y;2z = 2 x;2 y;Let {eq}f(x,y,z)=x^2y^2z^2 {/eq} and let S be the level surface defined by f(x,y,z) = 4 (a) Find an equation for the plane tangent to S at {eq}P_{0}(1,1,2)4 x4 y4 z4 =1 If x,y,z are nonzero, then we can consider Therefore, we have the following equations 1 1=2x2 2 1=2y2 3 1=2z2 4 x4 y4 z4 =1 Remember, we can only make this simplification if all the variables are nonzero!
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