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...

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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.**