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Ksivusya [100]
3 years ago
5

Jane needs open-topped boxes to store her excess inventory at year's end. She purchases large

Mathematics
1 answer:
Debora [2.8K]3 years ago
3 0

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

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