Question

An open top rectangular box is to be made from a rectangular piece of metal that is 3cm wide and 8cm long by cutting a square from each corner and then folding up the sides. Find the size of the corner squares that will give the box having the largest volume.

Answer 1

x = 0.67 cm

Explanation:

Let call " x " the length of the side of the square to cut from each corner

then the sides of the future box would be

L = 8 - 2x and D = 3 - 2x

The volume of the box is:

V = L*D*x

And such volume as function of x is

V(x) = ( 8 - 2x ) * ( 3 - 2x ) * x ⇒ V(x) = ( 24 - 16x - 6x + 4x²) * x

V(x) = 4x³ - 22x² + 24x

Taking derivatives on both sides of the equation we get:

V´(x) = 12x² - 44x + 24

Then V´(x) = 0 ⇒ 12x² - 44x + 24 = 0 ⇒ 3x² - 11x + 6 = 0

We got a second degree equation solving for x

x₁,₂ = [11 ± √ 121 - 72 ] / 6

x₁ = ( 11 + 7 ) / 6 x₁ = 3 we dismiss this solution since according to problem statement one side would become negative

Then

x₂ = ( 11 - 7 ) / 6 ⇒ x₂ = 4/6 ⇒ x₂ = 0.67 cm

As the second drivative is smaller than 0 then there is a maximun in that point

V´´(x) = 12x - 44 < 0

Sides of the box

L = 8 - 2x ⇒ L = 8 - 2*(0.67) ⇒ L = 8 - 1.34 ⇒ L = 6.66 cm

D = 3 - 2x ⇒ D = 3 - 2* (0.67) ⇒ D = 3 - 1.34 ⇒ D = 1.66 cm

Heigh = x = 0.67 cm

V(max) = 6.66*1.66*0.67

V(max) = 7.41 cm³