# Find the dimensions of the container with the greatest volume.The container, with a square base, vertical sides and an open top, is to be made from 1000sq ft of material.

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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.**

The surface material = 1000sq ft.

The container has has a square base . So the sides of the base be x and let h be the height.

Then the area of the base = x^2. sq feet.

The are of the four lateral surfaces = 4xh.

So the total surface = x^2+4xh = 1000.

So h = (1000-x^2)/4x

Volume of the container V(x) = x^2h = x^2(1000-x^2)/4x.

V(x) = 250x- (1/4)x^3.

To maximise V(x):

V'(x) = 0 gives V'(x) = 250 - (3/4)x^2.

So x^2= 250*4/3.

x = sqrt(1000/3) =18.26.

Also V''(x) = -(6/4)x . V''(18.26) = - (6/4)(18.26 < 0.

Therefore V(x) is maximum for x =

sqrt(1000/3) = 18.22674 ft and h = 9.1287 ft.