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V = x^2 [(2300-x^2)/4x]

V = 1/4 (2300x - x^3)

V' = 2300 - 3x^2 = 0

x = √(2300/3)

get the corresponding y-value then...

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get the volume...

x ≈ 27.689

y ≈ 13.844

V ≈ 10614.019 cc

The total surface area of the box with a square base and an open top = 2300

Area = A = area of square base + area of the four horizontal faces

Area of square base = x^2

Area of the the four horizontal sides = 4(xy)

where

x = length of the side of the square

y = depth of the box

Therefore,

2300 = x^2 + 4xy

Solving for "y",

y = (2300 - x^2)/4x

Volume of box = V = x^2y

and substituting y = (2300 - x^2)/4x,

V = x^2(2300 - x^2)/4x

V = x(2300 - x^2)/4

V = (1/4)(2300x - x^3)

dV/dx = (1/4)(2300 - 3x^2)

For maximum volume,

dV/dx = 0 = (1/4)(2300 - 3x^2)

3x^2 = 2300

x^2 = 2300/3

x = 27.69 cm.

Since x = 27.69

y = (2300 - x^2)/4x

y = (2300 - 766.74)/(4*27.69)

y = 13.84

Volume = (x^2)y

For maximum volume = (27.69)^2(13.84) = 10,614 cm^3

To CHECK if volume of box given the dimensions of "x" and "y" is maximum, get the 2nd derivative of the original V function. i.e.,

(d^2V/d^2x) = -(3/2)x

and since (d^2V/d^x) , 0, then the volume is maximum.

If side of square base = x and height of can = h

A = x^2 + 4xh

V = x^2h

h = V / x^2

A = x^2 + 4xV / x^2

A = x^2 + 4V / x

A = (x^3 + 4V) / x

2300x = x^3 + 4V

V = (2300x - x^3) / 4

For max vol dV/dx = 0

dV/dx = 2300/ 4 - 3x^2 / 4 = 0

3x^2 = 2300

x = Sqrt(2300/3)

therefore

Base area = 2300 / 3

h = (2300 - x^2) / 4x

h = (2300 - 2300/3) / 4*Sqrt(2300/3)

h = 13.84cm

V = 2300/3 * 13.84

V = 10610.6cm^3

let the base of square = x cm

height = h cm

surface area = x^2 +4xh

so x^2 + 4xh = 2300

4xh = 2300 - x^2

h = (2300 - x^2)/4x

volume = x^2*h

V = x^2(2300 - x^2)/4x

V = 575x - x^3/4

dV/dx = 575 - (3/4)x^2

In order V to be maximum

dV/dx = 0

575 - (3/4)x^2 = 0

(3/4)x^2 = 575

x^2 = (575*4)/3 = 2300/3

x = 27.69

h = (2300 - x^2)/4x

h = (2300 - 2300/3)/110.76

h = 13.84 cm

V = x^2*h

V = (2300/3)*13.84

V = 10613.54 cm^3