Question 1094699
 rectangular block of iron has to be cast so that its length is three times its width and its volume is a maximun,.
find the dimensions of the block if its total surface area is not to exceed 1800 square cm 
:
let L = the length of block
let x = the width
let h = the height
:
The surface area equation
2Lx + 2Lh + 2xh = 1800
simplify divide by 2
 xL + Lh + xh = 900 
we know that L = 3x
3x^2 + 3xh + xh = 900
3x^2 + 4xh = 900
4xh = 900 - 3x^2
h = {{{((900-3x^2))/(4x)}}} 

:
The volume equation
V = L * x * h
Replace L with 3x
V = 3x * x * h
V = 3x^2*h
Replace h with {{{((900-3x^2))/(4x)}}}
V = {{{(3x^2(900-3x^2))/(4x)}}}
cancel x
V = {{{(3x(900-3x^2))/(4)}}}
V = {{{((2700x - 9x^3))/4}}} or 675x - 2.25x^3
Plot this equation, volume vertical, width horizontal
{{{ graph( 300, 200, -6, 20, -1000, 5000, 675x-2.25x^3) }}} 
We can see max volume occurs when x=10 which is the width
:
Find the dimensions
L = 3(10)
L = 30 cm
w = 10 cm
h = {{{((900-3(10^2)))/(4*10)}}}
h = {{{(900-300)/40}}}
h = 15 cm is the height
:
30 * 10 * 15 = 4500 cu/cm is max volume
:
:
Check this by finding the surface area with these dimensions
2(30*10) + 2(30*15) + 2(10*15) =
600 + 900 + 300 = 1800 sq cm