【正文】 unctions: (1) (2) Solution. (1) ,soijkCritical points occur when ,. when(1) (2) (3) Using equations (2) and (3) to eliminate y and z from (1), we see thator ,giving , and .Hence we have three stationary points: , and . Since, and ,the Hessian matrix is At ,which has leading minors 0,And det .By the Leading Minor Test, then, is a local minimum. At ,which has leading minors 0,And det .By the Leading Minor Test, then, is also a local minimum.At , the Hessian isSince det, we can apply the leading minor test which tells us that this is a saddle point since the first leading minor is 0. An alternative method is as follows. In this case we consider the value of the expression，for arbitrarily small values of h, k and l. But for very small h, k and l, cubic terms and above are negligible in parison to quadratic and linear terms, so h, k and l are all positive, . However, if and and ,then .Hence close to ,both increases and decreases, so is a saddle point.(2) soij.Stationary points occur when ,. at .Let us classify this stationary point without considering the Leading Minor Test (in this case the Hessian has determinant 0 at so the test is not applicable).LetCompleting the square we see that So for any arbitrarily small values of h and k, that are not both 0, and we see that f has a local maximum at .2. Constrained Extrema and Lagrange MultipliersDefinition Let f and g be functions of n variables. An extreme value of f(x) subject to the condition g(x) = 0, is called a constrained extreme value and g(x) = 0 is called the constraint.Definition If is a function of n variables, the Lagrangian function of f subject to the constraint is the function of n+1 variableswhere is known as the Lagrange multiplier. The Lagrangian function of f subject to the k constraints , is the function with k Lagrange multipliers, Key Points.To find the extreme values of f subject to the constraint g(x) = 0: (1) calculate, remembering that it is a function of the n+1 variables and (2) find values of such that (you do not have to explicitly find the corresponding values of ): (3) evaluate f at these points to find the required extrema.Note that the equation is equivalent to the equations,and So, in the two variable case, we have Lagranian function and are solving the equations:, , and .With more than one constraint we solve the equation.Theorem Let and be a point on the curve C, with equation g(x,y) = 0, at which f restricted to C has a local that both and have continuous partial derivatives near to and that is not an end point of and that . Then there is some such that is a critical point of the Lagrangian Function.Proof. Sketch only. Since P is not an end point and ,has a tangent at with normal .If is not parallel to at , then it has nonzero projection along this tangent at .But then f increases and decreases away from along ,so is not an extremum. Henceand are parallel and there is some184。such that and the result follows.Example. Find the rectangular box with the largest volume that fits inside the ellipsoid ,given that it sides are parallel to the axes.Solution. Clearly the box will have the greatest volume if each of its corners touch the ellipse. Let one corner of the box be corner (x, y, z) in the positive octant, then the box has corners (177。x,177。y,177。z) and its volume is V= 8xyz.We want to maximize V given that . (Note that since the constraint surface is bounded a max/min does exist). The Lagrangian isand this has critical points when , . when (Note that will always be the constraint equation.) As we want to maximize V we can assume that so that .)Hence, eliminating , we getso that and But then so or ,which implies that and (they are all positive by assumption). So L has only one stationary point (for some value of , which we could work out if we wanted to). Since it is the only stationary point it must the required max and the max volume is. 附錄二：外文譯文 多元函數(shù)的極值1. 穩(wěn)定點 使并且. 對于任意一點有以下定義: (1)如果對于所有充分地接近時，則是一個局部極大值。(2)如果對于所有充分地接近時，則是一個局部極小值。(3)如果對于所有點成立，則是一個全局極大值（或絕對極大值）。(4) 如果對于所有點成立，則是一個全局極小值（或絕對極小值）。 (5) 局部極大（?。┲到y(tǒng)稱為局部極值；全局極大（?。┲到y(tǒng)稱為全局極值.定義 ，如果，并且對于任意奇異點都不存在，則稱是一個關(guān)鍵點或穩(wěn)定點.結(jié)論 , 則 一定是:(1)函數(shù)的一個關(guān)鍵點, 或者(2)函數(shù)的一個奇異點, 或者 (3)定義域的一個邊界點.結(jié)論 如果函數(shù)是一個在閉區(qū)間上的連續(xù)函數(shù)，則在區(qū)間上有邊界并且可以取到邊界值.定義 對于任一個關(guān)鍵點，當既不是局部極大值也不是局部極小值時，叫做函數(shù)的鞍點.結(jié)論 對于一個關(guān)鍵點是鞍點當且僅當任意小時，對于函數(shù)取正值和負值.定義