1. ## Minimisation Problem

Hi, I have gotten halfway through this problem but I have gotten stuck. The problem is...

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.

I have done the following.

Let the width = x, length = y and height = z; V (constant), V = xyz;

Cost = 5xy + 2xz + 2yz
Use V = xyz and sub z in for v/xy to get

Cost = 5xy + 2V/y + 2V/x

Minimizing this gives partial derivatives with respect to x and y

Costx = 5y - 2V/(x^2) *1
Costy = 5x - 2V/(y^2) *2

let y = 2V/5(x^2) from *1, sub this into *2

Giving...

x^3 = 2V/5

x = (2V/5)^1/3

Now do we make y = x and then solve for the z dimension? This is where I am lost.

Chris

2. Originally Posted by purakanui
Hi, I have gotten halfway through this problem but I have gotten stuck. The problem is...

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.

I have done the following.

Let the width = x, length = y and height = z; V (constant), V = xyz;

Cost = 5xy + 2xz + 2yz
Use V = xyz and sub z in for v/xy to get

Cost = 5xy + 2V/y + 2V/x

Minimizing this gives partial derivatives with respect to x and y

Costx = 5y - 2V/(x^2) *1
Costy = 5x - 2V/(y^2) *2

let y = 2V/5(x^2) from *1, sub this into *2

Giving...

x^3 = 2V/5

x = (2V/5)^1/3

Now do we make y = x and then solve for the z dimension? This is where I am lost.
Remember that you just said "let y = 2V/5(x^2) from *1". Solve for y from that.