The base of an aquarium with given volume V is made of slate and the sides are made of glass. If slate costs nine times as much (per unit area) as glass, find the dimensions of the aquarium that minimize the cost of the materials. (Let x, y, and z be the dimensions of the aquarium. Enter your answer in terms of V.) (x, y, z)

The dimensions of the aquarium that minimize the cost of the materials:

Step-by-step explanation:

Let x, y and z be the dimensions of aquarium .

Surface area of an aquarium = xy+2yz+2xz

Volume of aquarium V=

We are given that slate costs five times as much (per unit area) as glass

So, Cost function : C=9xy+2yz+2xz

Now we will use langrage multiplier to find the dimensions of the aquarium that minimize the cost of the materials.

So,

Multiply 1 ,2 and 3 by x,y and z respectively.

Now equate 4 and 5

9xy+2xz=9xy+2yz

x=y

Substitute y=x in 5 and 6 and equate them

Substitute the values in A

Hence,

The dimensions of the aquarium that minimize the cost of the materials:

b

step-by-step explanation:

the word superfluous means unnecessary

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)