A box with a square base is made using 1000 ft^2 of material.
Now we have to find the maximum possible volume of the box.
We know
that the volume of the box would be V=x^2*y, where x is the side of the square bottom and y is
the height.
The surface area of material used is x^2 + 4xy which is
equal to 1000 cm^2.
x^2 + 4xy =
1000
=> y = (1000 –
x^2)/4x
Substituting this in the expression for volume we get V =
x^2*(1000 – x^2)/4x
Now we have to maximize
V
V = x^2*(1000 – x^2)/4x
=> V =
x*(1000 – x^2) / 4
V’ = (1/4) [x* (-2x) + 1000 -
x^2]
=> (1/4) [- 2x^2 + 1000 –
x^2]
=> (1/4) [1000 –
3x^2]
Equate V’ to 0
=> (1/4)
[1000 – 3x^2] = 0
=> 3x^2 =
1000
=> x^2 = 1000 / 3
=>
x = sqrt 1000/3 [we don’t need the negative root]
=> x= 18.25
ft
We see that V’’ = -3x/2 which is negative, therefore V is maximum
for this value of x.
Now y = (1000 –
x^2)/4x
=> y = (1000 – 1000/3) / (4* sqrt(1000 /
3)
=> 2000 / 3*4 sqrt
1000/3
=> 9.128
ft
The required maximum volume is achieved with the
base having sides of 18.35 ft and the height 9.128 ft.
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