Draw a rectangle 7 x 12.
In the corners draw four squares of dimensions h x h.
It doesn't matter what the length of h is.
Dimensions of a new rectangle will be ( 7 - 2 h ) x ( 12 - 2 h )
The volume of the box will be:
Area of a new rectangle ∙ length h
V = ( 7 - 2 h ) ∙ ( 12 - h ) ∙ h
V = ( 84 - 24 h - 7 h + 2 h² ) ∙ h
V = ( 2 h² - 31 h + 84 ) ∙ h
V = 2 h³ - 31 h² + 84 h
The function has an exstrem (maximum or minimum) where the first derivative is zero.
In this case you need to find:
dV / dh = V′( h ) = 0
( 2 h³ - 31 h² + 84 h )′ = 2 ∙ 3 h² - 31 ∙ 2 h + 84 = 0
6 h² - 62 h + 84 = 0
The solutios are:
h = ( 31 -√457 ) / 6 ≈ 1.60374
and
h = ( 31 +√457 ) / 6 ≈ 8.7296
Now you have to second derivative test.
If V"( h ) < 0 then function has maximum at h.
If V" ( h ) > 0 then function has minimum at h.
V" ( h ) = 2 ∙ 6 h - 62
V" ( h ) = 12 h - 62
For
h = 1.60374
V" ( h ) = 12 ∙ 1.60374 - 62 = − 42.75512 < 0
For
h = 8.7296
V" ( h ) = 12 ∙ 8.7296 - 62 = 42.7552 > 0
So
For h = ( 31 -√457 ) / 6 ≈ 1.60374 volume has a maximum.
For h = ( 31 +√457 ) / 6 ≈ 8.7296 volume has a minimum.
By the way dimension h = 8.7296 are impossible because for h = 8.7296 dimension 7 - 2 h would have a negative value and the length cannot be negative.
A box (with no top) is to be constructed from a piece of cardboard of sides A and B by cutting out squares of length h from the corners and folding up the sides. Find the value of h that maximizes the volume of the box if
A = 7 and B = 12
1 answer
1 answer