Question

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)

Answer 1

Answer:

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

$x=y = \sqrt[3]{\frac{2V}{9}} , z = \sqrt[3]{\frac{81V}{4}}$

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= $Length \times Breadth \times Height =xyz----A$

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.

$\nabla C = \lamda \nabla V\\(\frac{\partial C}{\patial x},\frac{\partial C}{\partial y}, \frac{\partial C}{\partial z})=\lambda(\frac{\partial V}{\patial x},\frac{\partial V}{\partial y}, \frac{\partial V}{\partial z})\\(9y+2z,9x+2z,2y+2x)=\lambda(yz+zx+xy)$

So,$9y+2z=\lambda yz -----1\\9x+2z=\lambda zx -------2\\2y+2x=\lambda xy --------3$

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

$9xy+2xz=\lambda xyz ------4\\9xy+2zy=\lambda zxy -------5\\2yz+2xz=\lambda xyz --------6$

Now equate 4 and 5

9xy+2xz=9xy+2yz

x=y

Substitute y=x in 5 and 6 and equate them

$9x^2+2xz=2xz+2xz\\9x=2z\\\frac{9x}{2}=z$

Substitute the values in A

$Volume = xyz=(x)(x)(\frac{9x}{2})\\Volume V = \frac{9x}{2}x^3\\\sqrt[3]{\frac{2V}{9}}=x\\x=y=\sqrt[3]{\frac{2V}{9}}\\z=\frac{9(\sqrt[3]{\frac{2V}{9}})}{2}\\z=\sqrt[3]{\frac{81V}{4}}$

Hence,

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

$x=y = \sqrt[3]{\frac{2V}{9}} , z = \sqrt[3]{\frac{81V}{4}}$