Sec 4.5 Optimization II
In this section we would be given a word problem , where we would
be asked to mind the maximum or minimum value of a quantity.​ We
would have to formulate the function representing the quantity in
terms of the independent variable, and the interval in which the
variable is constrained to be in.​ Then find the absolute extrema
of the function over this interval.
Generally we follow the following two steps.
1. Draw a diagram​ describing the situation.
2. Choose the independent variable , and represent the function
in terms of this variable. Also determine the interval.
3. Find the absolute max/min
Q. By cutting away identical squares from each corner of a
rectangular piece of cardboard and folding up the resulting
flaps, an open box may be made. If the cardboard is ​17​ in. long
and ​16​ in. wide, find the dimensions of the box that will yield
the maximum volume. (Round your answers to two decimal places.)
Soln.
Let the length of the cuts be x . Then the height, width, length
of the box are x, 16 − 2x, 17 − 2x respectively. Hence the volume
V (x) = x(16 − 2x)(17 − 2x) = 272x − 66x2 + 4x3 . and x is in [0, 8] .
V ′(x) = 272 − 132x + 12x2 =0. Solving the equation we get the
solutions x= 8.253, 2.746.
We discard the 1st solution since it is not in the interval [0,8]
V(0) = 0, V(8)=0, V(2.746)=2.746(16 - 5.492)(17 - 5.492)= 332.06.
So the box will have maximum volume when x=2.746 and the
dimensions of the box are 2.746, 10.508, 11.508.
Q.
A rectangular box is to have a square base and a volume of
3​
40​ ft​ . If the material for the base costs $​ 0.35​ per square foot,
the material for the sides costs ​$0.05​ per square foot, and the
material for the top costs ​$0.15​ per square foot, determine the
dimensions of the box that can be constructed at minimum cost.
Soln. V = x2y = 40 . So y = 40/x2 .
C = 0.35 x2 + 4(0.05) xy + (0.15) x2 = 0.5 x2 + 0.2x(40/x2) = 0.5 x2 + 8/x . And
x is in (0, ∞) .
Please note how we wrote C(x) in terms of a alone( we eliminated
y)
The graph of C(x) in the interval (0, ∞) is
. From the graph we see that the absolute minima exists and at
that point f ′(x) = 0 . So we can find that point by solving f ′(x) = 0
.
f ′(x) = x − x82 . T he solution is x = 2 . So the dimensions when the cost is
minimum is 2,2,(40/4)=10.
Q. A book designer has decided that the pages of a book should
have 1-in. margins at the top and bottom and 12 -in margins
on the sides. She further stipulated that each page should have
an area of ​50​ in.2​ ​ (see the accompanying figure). Determine the
page dimensions that will result in the maximum printed area on
the page.
Soln. xy = 50. So y = 50/x . The printed area
50
P = (x − 1)(y − 2) = (x − 1)( 50
x − 2) = 50 − 2x − x + 2 . Since 1 ≤ x, 2 ≤ y , we
have 1 ≤ x ≤ 25 .
P ′(x) =− 2 + 50
x2 . P’(x) does not exist at 0 but that is outside our
interval. P’(x)=0 has solutions x= 5, -5. So the only critical
point in our interval is x = 5.
P(5)=32, P(1)=0, P(25)=0. So the maximim possible printed area is
32 in2 . and it is obtained when x=5, y=10.
Q. A Norman window has the shape of a rectangle surmounted by a
semicircle. Find the dimensions of a Norman window of perimeter
35​ ft that will admit the greatest possible amount of light.
(Round your answers to two decimal places.)
Soln. Perimeter P = πx + 2x + 2y = 35
⇒
y=
35−(π+2)x
2
.
Area = A(x) = (0.5)πx2 + 2xy = 0.5πx2 + 2x( 35−(π+2)x
) = 0.5πx2 + 35x − (π + 2)x2 .
2
35
x is in [0, π+2
].
A′(x) = πx + 35 − 2(π + 2)x = 35 − 4x − πx = 0 has solution x =
35
4+π
Clearly this is in our interval.
A(0)=0, A(
A(
35
π+4 )
35
π+2 )
35 2
35
= 0.5π( π+2
) + 2( π+2
)(
35 2
35
= 0.5π( π+4
) + 2( π+4
)(
35
35−(π+2)(π+2
)
)
2
35
35−(π+2)(π+4
)
)
2
35 2
= 0.5π( π+2
) =72.8
= 85.78
So the area of the window is maximized when x=
35
π+4
= 4.9, y =
35
35−(π+2)π+4
2
= 4.9
.