コンプリート！ f(x y z)=x^2 y^2 z^2 x^4 y^4 z^4=1 342181-F x y z x 2 y 2 z 2 x 4 y 4 z 4 1

X 4 y 4 z 4 = 1 BuySolution 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;Feb 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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F x y z x 2 y 2 z 2 x 4 y 4 z 4 1

F x y z x 2 y 2 z 2 x 4 y 4 z 4 1-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 3422 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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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) = 5Nov 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, insufficientApr 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?

Namely x 2= y From thisA 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 dA425 Calculate the limit of a

Z 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;(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 2x8 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 Share

Lagrange Multipliers Minimize F X Y X 2 Y 2 Subject To X 2y 5 0 Youtube

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Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, historyAnswer 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 By(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 2

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F ( x , y , z ) = x 2 y 2 z 2 ;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 respectFdS, 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 ,

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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;=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 isF(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 boundary

14 2 Limits And Continuity

Use Lagrange Multipliers To Find The Maximum And Minimum Values Of The Function Subject To The Given Constraint F X Y Z Xyz X 2 2y 2 3z 2 6 Homework Help And Answers Slader

The 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 5/6423 State the conditions for continuity of a function of two variables;In 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

Verify Stokes Theorem For V Zi Xj Yk Over The Hemispherical Surface X 2 Y 2 Z 2 1 And Z Gt 0 Mathematics Stack Exchange

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Incoming Term: f x y z x 2 y 2 z 2 x 4 y 4 z 4 1,