SOLUTION: a 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 no

Algebra ->  Rectangles -> SOLUTION: a 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 no      Log On


   

Question 1094699: a 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
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
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 = %28%28900-3x%5E2%29%29%2F%284x%29
:
The volume equation
V = L * x * h
Replace L with 3x
V = 3x * x * h
V = 3x^2*h
Replace h with %28%28900-3x%5E2%29%29%2F%284x%29
V = %283x%5E2%28900-3x%5E2%29%29%2F%284x%29
cancel x
V = %283x%28900-3x%5E2%29%29%2F%284%29
V = %28%282700x+-+9x%5E3%29%29%2F4 or 675x - 2.25x^3
Plot this equation, volume vertical, width horizontal
+graph%28+300%2C+200%2C+-6%2C+20%2C+-1000%2C+5000%2C+675x-2.25x%5E3%29+
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 = %28%28900-3%2810%5E2%29%29%29%2F%284%2A10%29
h = %28900-300%29%2F40
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