3.3. GLOBAL OPTIMIZATION
187
area of the square is A = s 2 =
20−x
4
2 . Therefore, the total area function is
A(x) =
√
3x 2
36
+
20 − x
4
2
.
Again, note that we are only considering this function on the restricted domain [0, 20] and
we seek its absolute minimum and absolute maximum.
Differentiating A(x), we have
A
′ (x) =
√
3x
18
+ 2
20 − x
4
−
1
4
=
√
3
18
x +
1
8
x −
5
2
.
Setting A ′ (x) = 0, it follows that x =
180
4
√
3+9
≈ 11.3007 is the only critical number of A,
and we note that this lies within the interval [0, 20].
Evaluating A at the critical number and endpoints, we see that
• A
180
4
√
3 + 9
=
√
3(
180
4
√
3+9
) 2
4
+
20 −
180
4
√
3+9
4
2
≈ 10.8741
• A(0) = 25
• A(20) =
√
3
36
(400) =
100
9
√
3 ≈ 19.2450
Thus, the absolute minimum occurs when x ≈ 11.3007 and results in the minimum area
of approximately 10.8741 square centimeters, while the absolute maximum occurs when
we invest all of the wire in the square (and none in the triangle), resulting in 25 square
centimeters of area. These results are confirmed by a plot of y = A(x) on the interval
[0, 20], as shown in Figure 3.20.
Activity 3.9.
A piece of cardboard that is 10 × 15 (each measured in inches) is being made into a box
without a top. To do so, squares are cut from each corner of the box and the remaining
sides are folded up. If the box needs to be at least 1 inch deep and no more than 3
inches deep, what is the maximum possible volume of the box? what is the minimum
volume? Justify your answers using calculus.
(a) Draw a labeled diagram that shows the given information. What variable
should we introduce to represent the choice we make in creating the box? Label
the diagram appropriately with the variable, and write a sentence to state what
the variable represents.
187
area of the square is A = s 2 =
20−x
4
2 . Therefore, the total area function is
A(x) =
√
3x 2
36
+
20 − x
4
2
.
Again, note that we are only considering this function on the restricted domain [0, 20] and
we seek its absolute minimum and absolute maximum.
Differentiating A(x), we have
A
′ (x) =
√
3x
18
+ 2
20 − x
4
−
1
4
=
√
3
18
x +
1
8
x −
5
2
.
Setting A ′ (x) = 0, it follows that x =
180
4
√
3+9
≈ 11.3007 is the only critical number of A,
and we note that this lies within the interval [0, 20].
Evaluating A at the critical number and endpoints, we see that
• A
180
4
√
3 + 9
=
√
3(
180
4
√
3+9
) 2
4
+
20 −
180
4
√
3+9
4
2
≈ 10.8741
• A(0) = 25
• A(20) =
√
3
36
(400) =
100
9
√
3 ≈ 19.2450
Thus, the absolute minimum occurs when x ≈ 11.3007 and results in the minimum area
of approximately 10.8741 square centimeters, while the absolute maximum occurs when
we invest all of the wire in the square (and none in the triangle), resulting in 25 square
centimeters of area. These results are confirmed by a plot of y = A(x) on the interval
[0, 20], as shown in Figure 3.20.
Activity 3.9.
A piece of cardboard that is 10 × 15 (each measured in inches) is being made into a box
without a top. To do so, squares are cut from each corner of the box and the remaining
sides are folded up. If the box needs to be at least 1 inch deep and no more than 3
inches deep, what is the maximum possible volume of the box? what is the minimum
volume? Justify your answers using calculus.
(a) Draw a labeled diagram that shows the given information. What variable
should we introduce to represent the choice we make in creating the box? Label
the diagram appropriately with the variable, and write a sentence to state what
the variable represents.