You are constructing a cardboard box from a piece of cardboard with the dimensions 2 m by 4 m. You then cut equal-size squares from each corner so you may fold the edges. What are the dimensions (in m) of the box with the largest volume

Dimensions:

l =3,15 m

w=1,15 m

x= 0,42 m   (height)

V(max) = 1,52 m³

Step-by-step explanation:

The cardboard is L = 4 m   and   W = 2 m

Let call x the length of the square to cut in each corner, then, volume of open box is:

For the side L       is  L - 2*x               l =  4 - 2*x

For the side W      is  W - 2*x             w=  2 - 2*x

The height              is  x

Volume of the open box, as function of x is:

V(x) = ( 4 -2x) * ( 2 - 2x) *x    ⇒  V(x) = ( 8 - 8x -4x + 4x²) *x

V(x) = ( 8 -12x + 4x² )*x     V(x) = 8x - 12x² + 4x³

V(x) = 8x - 12x² + 4x³

Taking derivatives on both sides of the equation

V´(x) = 8 - 24x + 12x²

V´(x) = 0       8  - 24x + 12x² = 0     reordering    12x² - 24x + 8  = 0

or    3x² - 6x + 2 = 0

A second degree equation. Solving for x

x₁,₂  = ( 6 ± √36 - 24 ) /6

x₁,₂  = ( 6 ± 3.46) /6

x₁  =  6 + 3,46 /6     x₁  = 1.58  we dismiss such solution because 1,58 * 2 = 3,15 and is bigger than 2 one of the side of the cardboard

x₂  =( 6 - 3,46 ) / 6

x₂  = 0,42 m

Dimensions of the open box

l  = 4 - 2*x     l  = 4 - 0,85      l  =  3,15 m

w = 2 -2*x    w  = 2 - 0,85     w = 1,15 m

x = 0,42 m

V(max) =3,15*1,15*0,42

V(max) = 1,52 m³

a right triangle. and in the future use this desmos.com to you

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this