A can of paint is 15 centimeters high and has a ddiameter of 13.6 cm. what is the volue of the can? Round to the nearest tenth.
2 answers:
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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
⇒ = - 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
⇒ = - λ .......(c)
<u>Step (iii):-</u>
Equating (a) and (b) equations
we get
cancel 'z' value we get x = 9y .......(d)
Equating (b) and (c) equations
we get
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