Respuesta :
Answer:
The dimensions of the aquarium that minimize the cost of the materials:
[tex]x=y=\sqrt[3]{\frac{2V}{5}}\\z=\sqrt[3]{\frac{25V}{4}}[/tex]
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=[tex]Length \times Breadth \times Height=xyz[/tex] ----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.
[tex]\nabla C =\lambda \nabla V[/tex]
[tex](\frac{\partial C}{\partial x},\frac{\partial C}{\partial y},\frac{\partial C}{\partial z})= \lambda (\frac{\partial V}{\partial x},\frac{\partial V}{\partial y},\frac{\partial V}{\partial z})[/tex]
[tex](5y+2z,5x+2z,2y+2x)=\lambda(yz,xz,xy)[/tex]
So,
[tex]5y+2z=\lambda yz[/tex] ----1
[tex]5x+2z=\lambda xz[/tex] -----2
[tex]2y+2x=\lambda xy[/tex] ----3
Multiply 1 ,2 and 3 by x,y and z respectively.
[tex]5xy+2xz=\lambda xyz[/tex] ----4
[tex]5xy+2yz=\lambda xyz[/tex] -----5
[tex]2yz+2xz=\lambda xyz[/tex] ----6
Now equate 4 and 5
5xy+2xz=5xy+2yz
x=y
Substitute y=x in 5 and 6 and equate them
[tex]5x(x)+2(x)z=2(x)z+2xz\\5x^2=2xz\\5x=2z\\\frac{5}{2}x=z[/tex]
Substitute the values in A
[tex]V = xyz = x \times x \times \frac{5}{2}xV=\frac{5}{2}x^3\\\sqrt[3]{\frac{2}{5}V}=x\\x=y=\sqrt[3]{\frac{2}{5}V}\\z=\frac{5}{2}x=\frac{5}{2}(\sqrt[3]{\frac{2}{5}})=\sqrt[3]{\frac{25V}{4}}[/tex]
Hence,
The dimensions of the aquarium that minimize the cost of the materials:
[tex]x=y=\sqrt[3]{\frac{2V}{5}}\\z=\sqrt[3]{\frac{25V}{4}}[/tex]