Multivariate 2 variate function differential calculus to find the best value problem

The length, width and height are a, b, and c

Then, from what is known: 18ab + 12bc + 12ac = 216 that is, 3ab + 2bc + 2ac = 36

So c = (36-3ab) (2a+2b).

So the partial derivative dc da=(36-3b 2) (a+b), dc db=(36-3a 2) (a+b).

v=abc=ab*(36-3ab) (2a+2b) to find the partial derivatives dv da and dv db, respectively

Then if and only if dv da=0 and dv db=0, v obtains the maximum, i.e., the maximum.

Solve a, b, then c can also know.

Using the method of differentiation, the expression of dz is found. Then, under the initial conditions, the value of z is found. Finally, substitution.

Hope. <>

12. e^(2yz)+x+y^2+z = 7/4 ①

x = y = 1 2, e z+z =1 , we get z = 0.

There are two ways to find the partial derivative:

Method 1: Find the partial derivative of x on both sides of the equation, note that z is a function of x,y, and obtain.

e (2yz)(2y z x)+1+ z x = 0, x = y = 1 2 , z = 0 substitution, get.

z/∂x+1+∂z/∂x = 0， ∂z/∂x = 1/2；

Finding the partial derivative of y on both sides of the equation, gets.

e (2yz)(2z+2y z y)+2y+ z y = 0, x = y = 1 2 , z = 0 substitution, get.

z/∂y+1+∂z/∂y = 0， ∂z/∂y = 1/2；

dz = 1/2)(dx+dy)

Method 2: Write f = e (2yz)+x+y 2+z-7 4.

fx = 1, fy = 2ze (2yz)+2y, fz = 2ye (2yz)+1, z x = fx fz = 1 [2ye (2yz)+1], z x = fy fz = 2ze (2yz)+2y] [2ye (2yz)+1], x = y = 1 2 , z = 0 substitution, z x = 1 2, z y = 1 2, dz = 1 2)(dx+dy).

<> pay attention to finding the partial derivative, to find the partial derivative of x, that is, to treat y as a constant, and then use the implicit function to find the derivative, and the same is true for y.

Here's how, please refer to:

<> according to the method of doing Hui lagrange and taking the town of rotten sons, it is indeed possible to get the pure brigade to answer.

f = xy+2yz+k(x^2+y^2+z^2-10)f'x = 0 : y+2kx = 0 i.e. y = 2kx scrambled width f'y = 0 : x+2z+2ky = 0 ②f'z = 0 :

2y+2kz = 0 i.e. y = kz f'k = 0 : x 2+y 2+z 2-10 = 0 to get z = 2x, substituting y gives y = 5x (2k) =2kx, gets k = 5 2,k = 5 2, y = 5x, together with z = 2x substitution gives x = 1,y = 5, z= 2,u = xy+2yz minimum u(1, -5, 2) =5 5, u(-1, 5, -2) =5 5；

k = 5 2, y = 5x, together with z = 2x substitution to get x = 1,y = 5, z= 2,u = xy+2yz the brightest value u(1, 5, 2) =5 5, u(-1, -5, -2) =5 5.

The multivariate function maximum-value problem in the postgraduate examination refers to solving the maximum or minimum value of the multivariate function under the given constraints. A common way to solve this type of problem is to use the Lagrange multiplier method. Here are the general steps:

1.Determine the objective function: First, determine the multivariate function that requires the solution maximum, usually expressed as f(x1, x2, .).xn)。

2.Determine Constraints: Determine the constraints, which are often expressed as g(x1, x2, ., .)xn) =c, where g(x1, x2, .).xn) denotes the constraint function, and c is a constant.

3.Build a Lagrangian function: Build a Lagrangian function l(x1, x2, ..)

xn, λf(x1, x2, .xn) +g(x1, x2, .xn) -c), where is the Lagrange multiplier.

4.Solve the partial derivative of the Lagrangian function: for the Lagrangian function l(x1, x2, ..)xn, find the partial derivatives of each variable and make it equal to the zero-range group: solve the system of equations to obtain the value of the variable and the value of the Lagrangian multiplier.

6.Test the maximum: substitute the obtained solution into the objective function f(x1, x2, .).xn) and constraints g(x1, x2, .).xn), the constraints are satisfied, and the maximum value is compared.

It is important to note that the Lagrangian multiplier method is applicable in the case of equality constraints. For inequality constraints, additional discussion processing is required.

The above are the general steps to solve the problem of the maximum value of multivariate functions in the graduate school entrance examination. The process of solving specific problems may be more complex, so it is recommended that we refer to relevant textbooks and exercises in the learning process to better understand and master the method.

Think of y as a constant, and then the normal derivative is fine.

Solution: xyz+ (x 2+y 2+z 2)= 2 Differentiation on both sides, resulting in d(xyz)+d( (x 2+y 2+z 2))=d( 2).

>yzdx+xzdy+xydz+(xdx+ydy+zdz)/√(x^2+y^2+z^2)=0

Therefore, the calculated differentiation is yzdx+xzdy+xydz+(xdx+ydy+zdz) (x 2+y 2+z 2)=0.

When finding the partial derivative for <>z for x, it is necessary to pay attention to finding the first derivative for u and v, and the same is true for y, so it is four. I wrote out the detailed process of finding the partial derivative of x, and the other two processes are similar, you can try it yourself, and you won't ask me again.