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3 years ago

I have no idea how to do this. Could i get some help? Thank you in advance.

Mathematics

1 answer:

fredd [130]3 years ago

8 0

Answer: #2. Small triangle-JK, KM, JM. Medium triangle-MK, KL, ML.

#3. Medium triangle-b, d, c. Original triangle-f, d, e.

Step-by-step explanation: For #2 the to letters put together are one side length. For example when you connect J and K it makes one side. So for the small triangle when you connect points JK, KM, and JM they create the small triangle. For the medium triangle when you connect points MK, KL, and ML they create the medium triangle. For #3 each letter is one side, and when you connect them they make a triangle. So for the medium triangle when you connect b, d, and c it creates the medium triangle. For the original triangle when you connect f, d, and e it creates the original triangle. I hope that this helped you a little.

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1) x=1
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3 years ago

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3 years ago

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:

Step 1:-

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)

Step( ii):

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)

Step (iii):-

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

y = 1

substitute y =1 in x = 9y

x = 9

substitute y =1 in 4z =9y

z = 9 /4

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

Conclusion:-

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

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