Respuesta :
Answer:
Box Dimensions:
L = 15.15 ul
W = 7.15 ul
h = x = 2.43 ul
V(max) = 263.22 cu
Step-by-step explanation:
We call x the length of the square to be cut in the corners then:
Are of the base of the box is:
(20 - 2*x) is the future length of the box and
(12 - 2*x) will be the width
The heigh is x then the volume of the box is:
V = ( 20 - 2*x )* ( 12 - 2*x ) * h
And the volume as a function of x is:
V(x) = ( 20 - 2*x) * ( 12 - 2*x ) * x or V(x) = (240 -40*x -24*x + 4*x²) * x
V(x) = 240*x - 64*x² + 4*x³
Taking derivatives on both sides of the equation we get:
V´(x) = 240 - 128*x + 12*x²
V´(x) = 0 240 - 128*x + 12*x² = 0 or 60 - 32*x + 3*x²
3*x² - 32*x + 60 = 0
Solving:
x₁,₂ = 32 ± √ (32)² - 4*3*60 ]/ 2*3
x₁,₂ = 32 ± √ 1024 - 720 )/6
x₁,₂ = ( 32 ± √ 304 )/6
x₁,₂ = ( 32 ± 17.44 )/6
x₁ = 8.23 ( we dismiss this solution because is not feasible 2*x > 12
x₂ = 2.43 u.l ( units of length)
Then
L = 20 - 2*x L = 20 - 4.85 L = 15.15 ul
W = 12 - 2*x W = 12 - 4.85 W = 7.15 ul
h = 2.43 ul
V = 2.43*7.15*15.15 cubic units
V = 263.22 cu
To see if when x = 2.43 function V has a maximum we go to the second derivative
V´´(x) = - 128 + (24)*2.43
V´´(x) = - 69.68 as V´´(x) < 0 then we have a maximum for V(x) in the point x = 2.43
