171k views
3 votes
The base of an aquarium with given volume V is made of slate and the sides are made of glass. If slate costs five 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.)

1 Answer

4 votes

Answer:

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


x=y=\sqrt[3]{(2V)/(5)}\\z=\sqrt[3]{(25V)/(4)}

Explanation:

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

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

Volume of aquarium V=
Length * Breadth * Height=xyz ----A

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

So, Cost function : C=5xy+2yz+2xz

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


\\abla C =\lambda \\abla V


((\partial C)/(\partial x),(\partial C)/(\partial y),(\partial C)/(\partial z))= \lambda ((\partial V)/(\partial x),(\partial V)/(\partial y),(\partial V)/(\partial z))


(5y+2z,5x+2z,2y+2x)=\lambda(yz,xz,xy)

So,


5y+2z=\lambda yz ----1


5x+2z=\lambda xz -----2


2y+2x=\lambda xy ----3

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


5xy+2xz=\lambda xyz ----4


5xy+2yz=\lambda xyz -----5


2yz+2xz=\lambda xyz ----6

Now equate 4 and 5

5xy+2xz=5xy+2yz

x=y

Substitute y=x in 5 and 6 and equate them


5x(x)+2(x)z=2(x)z+2xz\\5x^2=2xz\\5x=2z\\(5)/(2)x=z

Substitute the values in A


V = xyz = x * x * (5)/(2)xV=(5)/(2)x^3\\\sqrt[3]{(2)/(5)V}=x\\x=y=\sqrt[3]{(2)/(5)V}\\z=(5)/(2)x=(5)/(2)(\sqrt[3]{(2)/(5)})=\sqrt[3]{(25V)/(4)}

Hence,

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


x=y=\sqrt[3]{(2V)/(5)}\\z=\sqrt[3]{(25V)/(4)}

User Tonytonov
by
4.8k points