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

Answer:

x  = 0.67 cm

Step-by-step 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