A rectangular box with a square base and open top is to be made from 1200 cm^2 of material. Find the dimensions of the box with the largest volume.

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The box has a square base and has a open top. Let the side of the square base be S and the height of the box be H.

The surface area of the box is S^2 + 4*S*H.

S^2 + 4*S*H = 1200

=> H = 300/S - S/4

The volume of the box is V = S^2*H = S^2*(300/S - S/4)

=> V = 300*S - S^3/4

To maximize V, solve `(dV)/(dS) = 0`

=> 300 - (3*S^2)/4 = 0

=> (3*S^2)/4 = 300

=> S^2 = 400

=> S = 20

`(d^2V)/(dV^2) = -3/4*(2*S)` which is negative for S = 20. This shows that the maximum volume of the box is when S = 20.

H = 300/S - S/4 = 15 - 5 = 10

The maximum volume is 4000 cm^3.

**The dimensions for the maximum volume are 20x20x10 which gives a volume of 4000 cm^3.**

let x be side of the square base and h the height of the box. The surface are of the box = x^2+4xh

This gives x^2 + 4xh = 1200 cm

h = (1200-x^2)/4x

The volume of the box v= x^2*(1200-x^2)/4x

v=300x-(x^3)/4

for max volume: d'v=0

d'v = 300-(3x^2)/4 = 0

=> 300-(3x^2)/4 = 0

=> x^2 = 200

x = 20 cm

h = (1200-x^2)/4x = (1200-20^2)/(4*20) = 10cm

surface area = 20^2+4*20*10 = 1200cm2

Volume = 20*20*10 = 4000 cm3

**2.1) Dimensions of the box with max volume are 20cm*20cm*10cm**

to confirm that the box has max volume we find d"v

d'v = 300-(3x^2)/4

d"v = -6x/4 = -3x/2

d"v = -3*20/2 = -30 for x=20

**2.2) as d"v is negative therefore the volume is maximum.**

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