Maximizing volume given a surface area constraint
Math 8
Department of Mathematics
Dartmouth College
Maximizing volume given a surface area constraint – p.1/9
Maximizing wih a constraint
We wish to solve the following problem:
Find the maximum volume of a rectangular box that has a surface area of 70cm 2 .
Maximizing volume given a surface area constraint – p.2/9
Maximizing wih a constraint
We wish to solve the following problem:
Find the maximum volume of a rectangular box that has a surface area of 70cm 2 .
• We begin by translating this into a set of equations:
V = lwh
SA = 2lw + 2wh + 2lh
Maximizing volume given a surface area constraint – p.2/9
Maximizing wih a constraint
We wish to solve the following problem:
Find the maximum volume of a rectangular box that has a surface area of 70cm 2 .
• We begin by translating this into a set of equations:
V = lwh
SA = 2lw + 2wh + 2lh
• Solving the surface area equation for h yields
35 − lw
h=
l+w
Maximizing volume given a surface area constraint – p.2/9
Defining V in terms of l and w
So far, we have
35 − lw
h=
l+w
• Plugging this into the volume equation yields
35lw − l2 w2
35 − lw
V = lw =
l+w l+w
Maximizing volume given a surface area constraint – p.3/9
Defining V in terms of l and w
So far, we have
35 − lw
h=
l+w
• Plugging this into the volume equation yields
35lw − l2 w2
35 − lw
V = lw =
l+w l+w
• We will now maximize this function for positive values of l and w - negative values
do not make sense physically.
Maximizing volume given a surface area constraint – p.3/9
Defining V in terms of l and w
So far, we have
35 − lw
h=
l+w
• Plugging this into the volume equation yields
35lw − l2 w2
35 − lw
V = lw =
l+w l+w
• We will now maximize this function for positive values of l and w - negative values
do not make sense physically.
To do this, we follow our maximization/minimization procedure.
1. Find the critical points of V with l ≥ 0 and w ≥ 0
2. Test critical points and boundary points to find maximum
Maximizing volume given a surface area constraint – p.3/9
Calculating Vl
First, calculate the partial derivative of V with respect to l:
∂ ∂ 35lw − l2 w2
V =
∂l ∂l l+w
Maximizing volume given a surface area constraint – p.4/9
Calculating Vl
First, calculate the partial derivative of V with respect to l:
∂ ∂ 35lw − l2 w2
V =
∂l ∂l l+w
(35w − 2lw 2 )(l + w) − 35lw + l2 w2 35lw − 2l2 w2 + 35w 2 − 2lw 3 − 35lw + l2 w2
= =
(l + w)2 (l + w)2
Maximizing volume given a surface area constraint – p.4/9
Calculating Vl
First, calculate the partial derivative of V with respect to l:
∂ ∂ 35lw − l2 w2
V =
∂l ∂l l+w
(35w − 2lw 2 )(l + w) − 35lw + l2 w2 35lw − 2l2 w2 + 35w 2 − 2lw 3 − 35lw + l2 w2
= =
(l + w)2 (l + w)2
Maximizing volume given a surface area constraint – p.4/9
Calculating Vl
First, calculate the partial derivative of V with respect to l:
∂ ∂ 35lw − l2 w2
V =
∂l ∂l l+w
(35w − 2lw 2 )(l + w) − 35lw + l2 w2 35lw − 2l2 w2 + 35w 2 − 2lw 3 − 35lw + l2 w2
= =
(l + w)2 (l + w)2
−l2 w2 + 35w 2 − 2lw 3
=
(l + w)2
Maximizing volume given a surface area constraint – p.4/9
Calculating Vl
First, calculate the partial derivative of V with respect to l:
∂ ∂ 35lw − l2 w2
V =
∂l ∂l l+w
(35w − 2lw 2 )(l + w) − 35lw + l2 w2 35lw − 2l2 w2 + 35w 2 − 2lw 3 − 35lw + l2 w2
= =
(l + w)2 (l + w)2
−l2 w2 + 35w 2 − 2lw 3
=
(l + w)2
w2 2
= (35 − 2lw − l )
(l + w)2
Maximizing volume given a surface area constraint – p.4/9
Calculating Vw
Second, calculate the partial derivative of V with respect to w:
∂ ∂ 35lw − l2 w2
V =
∂w ∂w l+w
Maximizing volume given a surface area constraint – p.5/9
Calculating Vw
Second, calculate the partial derivative of V with respect to w:
∂ ∂ 35lw − l2 w2
V =
∂w ∂w l+w
(35l − 2l2 w)(l + w) − 35lw + l2 w2 35l2 − 2l3 w + 35lw − 2l2 w2 − 35lw + l2 w2
= =
(l + w)2 (l + w)2
Maximizing volume given a surface area constraint – p.5/9
Calculating Vw
Second, calculate the partial derivative of V with respect to w:
∂ ∂ 35lw − l2 w2
V =
∂w ∂w l+w
(35l − 2l2 w)(l + w) − 35lw + l2 w2 35l2 − 2l3 w + 35lw − 2l2 w2 − 35lw + l2 w2
= =
(l + w)2 (l + w)2
Maximizing volume given a surface area constraint – p.5/9
Calculating Vw
Second, calculate the partial derivative of V with respect to w:
∂ ∂ 35lw − l2 w2
V =
∂w ∂w l+w
(35l − 2l2 w)(l + w) − 35lw + l2 w2 35l2 − 2l3 w + 35lw − 2l2 w2 − 35lw + l2 w2
= =
(l + w)2 (l + w)2
35l2 − l2 w2 − 2l3 w
=
(l + w)2
Maximizing volume given a surface area constraint – p.5/9
Calculating Vw
Second, calculate the partial derivative of V with respect to w:
∂ ∂ 35lw − l2 w2
V =
∂w ∂w l+w
(35l − 2l2 w)(l + w) − 35lw + l2 w2 35l2 − 2l3 w + 35lw − 2l2 w2 − 35lw + l2 w2
= =
(l + w)2 (l + w)2
35l2 − l2 w2 − 2l3 w
=
(l + w)2
l2 2
= (35 − 2lw − w )
(l + w)2
Maximizing volume given a surface area constraint – p.5/9
Find critical points
From the previous computations, we have
w2 l2
2
∇V (l, w) = 2
(35 − 2lw − l ), 2
(35 − 2lw − w 2 )
(l + w) (l + w)
Maximizing volume given a surface area constraint – p.6/9
Find critical points
From the previous computations, we have
w2 l2
2
∇V (l, w) = 2
(35 − 2lw − l ), 2
(35 − 2lw − w 2 )
(l + w) (l + w)
Assuming, for physical reasons, that l and w are nonzero, we need to solve the system
35 − 2lw − l2 = 0
35 − 2lw − w 2 = 0
Maximizing volume given a surface area constraint – p.6/9
Find critical points
From the previous computations, we have
w2 l2
2
∇V (l, w) = 2
(35 − 2lw − l ), 2
(35 − 2lw − w 2 )
(l + w) (l + w)
Assuming, for physical reasons, that l and w are nonzero, we need to solve the system
35 − 2lw − l2 = 0
35 − 2lw − w 2 = 0
• Subtracting the second equation from the first yields:
(35 − 2lw − l2 ) − (35 − 2lw − w 2 ) = 0
−l2 + w2 = 0
w = ±l
Maximizing volume given a surface area constraint – p.6/9
Find critical points
From the previous computations, we have
w2 l2
2
∇V (l, w) = 2
(35 − 2lw − l ), 2
(35 − 2lw − w 2 )
(l + w) (l + w)
Assuming, for physical reasons, that l and w are nonzero, we need to solve the system
35 − 2lw − l2 = 0
35 − 2lw − w 2 = 0
• Subtracting the second equation from the first yields:
(35 − 2lw − l2 ) − (35 − 2lw − w 2 ) = 0
−l2 + w2 = 0
w = ±l
• Again, to be physically reasonable, both l and w are positive so, l = w
Maximizing volume given a surface area constraint – p.6/9
Solving for critical points
So far, we have that l = w.
Plugging this into the first equation 35 − 2lw − l 2 = 0, we get
35 − 2l2 − l2 = 0
Maximizing volume given a surface area constraint – p.7/9
Solving for critical points
So far, we have that l = w.
Plugging this into the first equation 35 − 2lw − l 2 = 0, we get
35 − 2l2 − l2 = 0
Or,
35
l2 =
3
Maximizing volume given a surface area constraint – p.7/9
Solving for critical points
So far, we have that l = w.
Plugging this into the first equation 35 − 2lw − l 2 = 0, we get
35 − 2l2 − l2 = 0
Or,
35
l2 =
3
Or, again noting the l must be positive:
1
35 2
l=
3
Maximizing volume given a surface area constraint – p.7/9
Solving for critical points
So far, we have that l = w.
Plugging this into the first equation 35 − 2lw − l 2 = 0, we get
35 − 2l2 − l2 = 0
Or,
35
l2 =
3
Or, again noting the l must be positive:
1
35 2
l=
3
35−lw
Putting this together with the fact that w = l and h = l+w
, we have that the only
critical point in our domain is
1 1 1 1 1 !
35 − 35
35 2 35 2
3 35 2 35 2 35 2
, , 1 = , ,
3 3 2 335 2
3 3 3
Maximizing volume given a surface area constraint – p.7/9
Finding the absolute maximum
• The boundary of our domain are the lines l = 0 and w = 0 for l, w ≥ 0. Noticing
that on these lines, the volume and the surface area must be zero, we can ignore
these as physically unreasonable solutions.
Maximizing volume given a surface area constraint – p.8/9
Finding the absolute maximum
• The boundary of our domain are the lines l = 0 and w = 0 for l, w ≥ 0. Noticing
that on these lines, the volume and the surface area must be zero, we can ignore
these as physically unreasonable solutions.
• Thus our critical point is the only test point. The Volume at this point is
1 !3
35 2
V =
3
Maximizing volume given a surface area constraint – p.8/9
Finding the absolute maximum
• The boundary of our domain are the lines l = 0 and w = 0 for l, w ≥ 0. Noticing
that on these lines, the volume and the surface area must be zero, we can ignore
these as physically unreasonable solutions.
• Thus our critical point is the only test point. The Volume at this point is
1 !3
35 2
V =
3
• The only remaining question is, is this a maximum or not?
Maximizing volume given a surface area constraint – p.8/9
Finding the absolute maximum
• The boundary of our domain are the lines l = 0 and w = 0 for l, w ≥ 0. Noticing
that on these lines, the volume and the surface area must be zero, we can ignore
these as physically unreasonable solutions.
• Thus our critical point is the only test point. The Volume at this point is
1 !3
35 2
V =
3
• The only remaining question is, is this a maximum or not?
• To answer this we make two observations:
Maximizing volume given a surface area constraint – p.8/9
Finding the absolute maximum
• The boundary of our domain are the lines l = 0 and w = 0 for l, w ≥ 0. Noticing
that on these lines, the volume and the surface area must be zero, we can ignore
these as physically unreasonable solutions.
• Thus our critical point is the only test point. The Volume at this point is
1 !3
35 2
V =
3
• The only remaining question is, is this a maximum or not?
• To answer this we make two observations:
• 35−w
For a fixed l (say l = 1), we have h = 1+w
. As w → 0, we see that h → 35 and
the volume tends to zero.
Maximizing volume given a surface area constraint – p.8/9
Finding the absolute maximum
• The boundary of our domain are the lines l = 0 and w = 0 for l, w ≥ 0. Noticing
that on these lines, the volume and the surface area must be zero, we can ignore
these as physically unreasonable solutions.
• Thus our critical point is the only test point. The Volume at this point is
1 !3
35 2
V =
3
• The only remaining question is, is this a maximum or not?
• To answer this we make two observations:
• 35−w
For a fixed l (say l = 1), we have h = 1+w
. As w → 0, we see that h → 35 and
the volume tends to zero.
• As l and w tend towards ∞, we see that h eventually becomes negative, an
unreasonable physical assumption.
Maximizing volume given a surface area constraint – p.8/9
Conclusion
We conclude that the critical point is a maximum. Thus the absolute maximum volume
occurs at the point
1 1 1 !
35 2 35 2 35 2
, ,
3 3 3
where the volume is
3
35 2
Maximizing volume given a surface area constraint – p.9/9
