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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
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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