Answer:
x = 1,6 in ( the side of the corner squares )
V(max) = 67,58 in³
Step-by-step explanation:
The cardboard is:
L = 12 in w = 8 in
Let´s call "x" the side of the square from the corner:
Then the sides of the base of the open box are:
( L - 2*x ) and ( w - 2*x ) and x is the height
( 12 - 2*x ) and ( w - 2*x )
V(ob) = (L - 2*x ) * ( - 2*x ) * x
Wich is a function of x
V(x) = [( 12 - 2*x ) * ( 8 - 2*x ) ]*x
V(x) = ( 96 - 24*x - 16*x + 4*x²) * x
V(x) = 96*x - 40*x² + 4*x³
Tacking derivatives on both sides of the equation
V´(x) = 12*x² - 80*x + 96
V´(x) = 0 12*x² - 80*x + 96 = 0
Solving for x
x₁,₂ = 80 ± √ 6400 - 4608 / 24
x₁,₂ = 80 ± 42,33 / 24
x₁ = 5,10 We dismiss this solution since 2*x becomes 2*5,10 = 10,20
ant this value is bigger than 8 inches
x₂ = 1,60 in
Therefore dimensions of the box
a = 12 - 2*x ; a = 12 - 3,20 ; a = 8,8 in
b = 8 - 2*x ; b = 8 - 3,20 ; b = 4,80 in
And the volume of the open box is:
V(max) = 8,8*4,8*1,6
V(max) = 67,58 in³
How do we Know that is the maximun value for V?
We find V´´(x) = 24*x - 80 for x = 1,6 is negative ( V´´(x) = - 41,6 therefore V (x) has a local maximun for a value of x = 1,6
