CALCULUS 3 (LAGRANGE MULTIPLIERS) - Essay Example

Langrage Multipliers Finding the minima or maxima of a function is a common problem in calculus. face the problem of finding a closed form of the objective function (Courant, 36). This paper discusses how the method of langrage multipliers can help solve these problems without explicitly solving the conditions and using them to remove extra variables. Langrage multipliers are helpful when some of the variables in a restraint are redundant (Stewart, 65). For instance, it is not enough to ask, “How can one minimize the gold needed to make a cup?

” instead, one should ask, “How can one minimize the gold while making sure the cup will hold half a liter of tea.”Application of Langrage multipliersIt is possible to find the minima or maxima of a given function f of several variables given that the variables have restrictions. Langrage multipliers help us minimize or maximize functions with constraints that are points on a definite surface (Stewart, 67). Finding critical points of a function f (w, x, y) on a level surface or subject to the constraint g (w, x, y) = M.

Solving this problem requires a solving the following series of simultaneous equations.∆ f (w, x, y) = λ ∆g (w, x, y)g (w, x, y) = MRecalling that ∆ f and ∆g are vectors, we can write the above equations as follows: fw (w, x, y) = λ gw (w, x, y) fx (w, x, y) = λ gx (w, x, y) fy (w, x, y) = λ gy (w, x, y) g (w, x, y) = Mλ is a dummy variable known as a langrage multiplier. The most important variable in the equation are w, x, and y. After solving for the critical values, you insert them into the function to find the maxima and the minima.

The critical points where the function is greatest are maxima while where it is smallest are minima (Stewart, 72).Solving the system equations can be a difficult task. In order to solve for the critical points in an easier way, one should consider the following tricks; Solve for λ in terms of the variables w, x, and y to eliminate it from the equations. Solve any of the variables in terms of the other variables. Consider both the positive and negative square roots whenever using a square root (Courant, 38).

ExampleUse the langrage multiplier to find the critical points of the function f (w, x) = wx on the curve 3w2 + x2 = 6SolutionFrom the above equations;∆f = (w, x) and ∆g = (6w,2x)Therefore, we should solvex = 6λw (1)w = 2λx (2)3w2 + x2 = 6 (3)Inserting equation 1 into 2 givesX = 6λ (2λx) = 12λ2x (4)If x were zero, then w would be zero, which is impossible as 3w2 + x2 = 6Dividing x all through from equation (4) we get 12λ2 = 1Inserting equation 1 into equation 36 = 3w2 + (6λw) 2 = 3w2 + 36λ2w2 = 3w2 + 3(12λ2) w2 = 3w2 + 3w2Therefore w = ±1 and x = ±√3 by equation 3.

There are four critical points: (1,-√3), (1, √3), (-1, -√3) and (-1, √3).Work CitedCourant, Richard, and E. J. McShane.Differential and integral calculus. [2nd Ed. New York: Interscience, 2007. Print.Stewart, James. Calculus. 5th ed. Belmont, CA: Thomson Brooks/Cole, 2003. Print.