Maximum/Minimum Investigation Worksheet Solutions
1. Let x be the length of a square to be cut out. Then V(x) = x∙(11 - 2x)(8.5 - 2x) =
93.5x – 39x2 + 4x3. This means V’(x) = 12x2 – 78x + 93.5, and when we set V’(x) = 0 and
solve, we get that x = 1.585 inches. V(1.585) ≈ 66.15 in3, which is the maximum volume.
2. Let x be the length of the rectangle. Use similar triangles to get that the width is
30 – ¾ x, so A(x) = x∙(30 – ¾ x). Being quadratic, A(x) has roots at 0 and 40, so the
maximum must occur at the average of the roots which is 20. If the length is 20’,
the width will be 15’ and the maximum area is 300 ft2.
3. Let x be the width of the square that we cut out. Therefore V(x) = x∙(1 - x)(1 - x) =
x – 2x2 + x3. This means V’(x) = 3x2 – 4x + 1 = (3x-1)(x-1). So if x = 1/3 m,
V’(x) = 0, and V(1/3) = 4/27 m3, which is the maximum volume of the box.
4. By the definition, S = w∙l2, and since w2 + l2 = 302, l2 = 302 – w2. Substituting, we
get S(w) = w∙(302 – w2) = 900w – w3. S’(w) = 900 – 3w2, and solving 900 – 3w2 = 0
yields w = 10 ≈ 17.32 cm, which is the width needed for the maximum strength.
Thus, the corresponding length is approximately 24.49 cm.
5. a) Let x be the length (and the width) of the suitcase. V = x∙x∙(158 - 2x) = 158x2 – 2x3.
V’(x) = 316x – 6x2 = 0 x = 52.67 cm. Therefore the dimensions of the suitcase would
be 52.67 cm X 52.67 cm X 52.67 cm, giving a maximum volume of 146085 cm3.
b) Now the volume is V = x∙2x∙(158 – 3x). So V’(x) = 632x – 18x2, and when V’(x) = 0,
x = 35.11, so the new dimensions are 35.11 cm X 70.22 cm X 52.66 cm.
6. Let x be the radius of the circle. Then diameter is 2x, the circumference of the circle
is 2πx, and so the two straight segments must have length of (400 – 2πx)/2 = 200 – πx.
a) A(x) = (200 – πx)(2x). This is a quadratic with roots of 0 and 200/π. Therefore if
the radius of the circle is 100/π, the dimensions of the field will be approximately
63.66 m X 100 m and the maximum area of the rectangular field will be 6366 m2.
b) A(x) = (200 – πx)(2x) + πx2 = 400x – πx2 = x(400 – πx). This yields a maximum when
x = 200/π ≈ 63.33, meaning the diameter is 400/π and if the diameter is 400/π, the
track is a circle. (There is no rectangular field in the middle.)
7. Let x be the distance she runs. Then 150 – x is the remaining part of the beach and
by the Pythagorean theorem, she must swim to rescue the swimmer.
The time it takes to run (since d = r∙t, t = d/r) is x/3, and the time it takes to swim is
/1, so T(x) = x/3 + . Since we can’t take T’(x), we’ll
find the minimum time by graphing, and get that if she runs 142.0 m, her minimum time
will be 107.9 seconds.
8. Let X be the length of cable on ground. Then 180 – x is the remaining distance on land,
And by the Pythagorean theorem, is the length of cable that must lie
Under water. The cost will be C(x) = 30000x + 50000. Once again we
must use a calculator to get that if 135 feet of cable is laid on land, the minimum cost
for the cable is $7,800,000.