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ryzh [129]
3 years ago
8

Solve the system of equations and choose the correct ordered pair. 3x + 4y = 38 5x - 5y = -30

Mathematics
1 answer:
faust18 [17]3 years ago
3 0
3x + 4y = 38 ( equation 1) ----> ×5 5x - 5y = -30 ( equation 2) ----> ×3 When you multiply eqn 1 by 5 you get, 15x + 20y = 190 And when you multiply eqn 2 by 3, you get, 15x - 15y = -90 Then you solve both equations by subtracting eqn 2 from eqn 1 15x + 20y = 190 15x - 15y = 90 Then 15x - 15x gets cancelled and 20y - (-15y) gives 35y and 190 - (-90) gives 280. So that gives 35y = 280 y = 8 And when you replace y = 8 in any of the two equations, x = 2
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Use Lagrange multipliers to find the volume of the largest rectangular box in the first octant with three faces in the coordinat
tensa zangetsu [6.8K]

Answer:

The volume of the largest rectangular box (V) = 81/4

Step-by-step explanation:

<u>Step 1</u>:-

Given volume of the largest rectangular box in the first octant

V = l b h

let (x ,y, z) be the one vertex in the given plane

V = f(x, y, z) = x y z

given plane.   Ф (x, y, z) = x + 9y + 4z = 27 ........(1)

<u>Step( ii):</u>

By using  Lagrange multipliers

Suppose it is required to find the extreme for the function f(x, y, z) subject to the condition Ф (x, y, z) =0

Form Lagrange function F(x, y, z) = f(x, y, z) + λ  Ф (x, y, z) where λ is called the Lagrange multipliers

F(x, y, z) = x y z+ λ ( x + 9y + 4z - 27) ......(2)

Obtain the equations are δ F / δ x = 0

                                  δ f / δ x +λ  δ Ф  / δ x =0

                           ⇒ y z + λ ( 1) = 0

                          ⇒ y z = - λ ......(a)

Obtain the equations are δ F / δ y = 0

                            δ f / δ x +λ  δ Ф  / δ x =0

                        ⇒ x z + λ ( 9) = 0

                        ⇒ \frac{xz}{9} = - λ .......(b)

Obtain the equations are δ F / δ z = 0

                                  δ f / δ x +λ  δ Ф  / δ z =0

                                   x y + λ ( 4) = 0

                                    ⇒ \frac{xy}{4} = - λ .......(c)

<u>Step (iii):-</u>

Equating (a) and (b) equations

we get         y z = \frac{xz}{9}

cancel 'z' value we get  x = 9y  .......(d)

Equating (b) and (c) equations

we get     \frac{xy}{4}  = \frac{xz}{9}

cancel 'x' value on both sides , we get

              4z =9y  ......(e)

substitute (d) (e) values in equation (1)  we get

x + 9y + 4z -27 = 0

9y + 9y + 9y -27 =0

27y -27 =0

<u> y = 1</u>

substitute y =1 in x = 9y

<u>x = 9</u>

substitute y =1 in 4z =9y

<u>z = 9 /4</u>

therefore the dimensions are x =9 , y=1 and z = 9 /4

<u>Conclusion</u>:-

The largest volume of the rectangular box V = x y z

substitute x =9 , y=1 and z = 9 /4

V = 9(1)(9/4) =81/4

The largest volume of the rectangular box (V) = 81/4

Verification :-

Given plane x + 9y + 4z = 27

substitute x =9 , y=1 and z = 9 /4

              9 + 9 + 4(9/4) = 27

           27 =27

so satisfied equation

     

3 0
3 years ago
Https://aatmagyan.page.link/jJM5pmd5qbgW27KH7​
krek1111 [17]

what is this?

?

please help me

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3 years ago
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