An open rectangular box with square base is to be made from 48 ft^2 of material. What dimensions will result in a box with the largest possible volume?

bear with me.. solving

Surface area is area of the bottom: x^2 + 4 sides: 4*xy Surface area = 48ft^2 = x^2 + 4xy Volume is x^2*y: V = x^2 * y

Solve for y in the surface area: 48 = x^2 + 4xy 48-x^2 = 4xy (48-x^2)/4x = y

y = (48 - x^2) / 4x V = x^2 * y = x^2 * (48 - x^2) / 4x = x ( 48-x^2) / 4

Simplify a bit: V = 48x/4 - x^3/4 = 12x - x^3/4 Take the derivative and set it equal to 0: V ' = 12 - 3x^2 / 4 = 0

Solve the equation: 12 - 3x^2 / 4 = 0 -3x^2/4 = -12 x^2/4 = 4 x^2 = 4*4 = 16

x = + or - 4, but in this case it's a plus because the edge of the box cannot be negative.. anyway let's double check if it's min or max. To do so, take the 2nd derivative: V" = -6x / 4 = - 3x / 2. Which is positive if x = -4 and negative if x = 4. Maximum occurs when 2nd derivative is negative therefore x = 4 is max

Okay so now that you have x = 4, find y: y = (48 - x^2) / 4x y = (48 - 16) / 16 = 32/16 = 2 So dimensions are: The sides of the base must be equal to 4, and the height of the box must be 2. In which case the volume = area of the base * height: V = 4^2 * 2 = 16 * 2 = 32

Hope this helped =)

by the way units are: x = 4ft y = 2ft V = 32ft^3

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