- 1.
- Maximize
.
- 2.
- Cut out a square of side
from each corner.
- 3.
- Maximize
, getting a maximum area of 20,000 sq. yd.
- 4.
- 25 feet by 10 feet
- 5.
- Maximize
.
- 7.
- Maximize
to get
.
- 8.
- (a) The triangle will have a maximum altitude when the vertex is on
the diameter perpendicular to the chord. That makes the triangle isosceles.
(b) If the triangle has maximum area, it is isosceles. Let the base chord be a distance from the center of the circle. Then the altitude of the triangle is and the base is . The area is then , which has a critical point ar . It follows that each side of the triangle is , so it is equilateral. The area is .

- 9.
- Maximize .
- 10.
- If the vertex of the rectangle in the first quadrant has
coordinates
, then the area is
and has its
maximum at
. The maximum area is
.
- 12.
- If the trapezoid has altitude , then its area is
. is maximum at
, and its
maximum value is
.
- 13.
- Maximize
.
- 18.
- Let the can have height and base radius , with given
volume . The amount of material used is the same as the total area, which
is
. Using
, we eliminate and get
, which has its minimum at
. It follows that .
- 20.
- If is the rent, then the profit is
, which has maximum at .
- 23.
- We set and find that the maximum sustainable removal
occurs at . That is, fishermen may remove up to 1200 fish each year
without depleting the fish population.
- 24.
- Let be the angle of elevation to the bottom of the
tapestry and let be the angle subtended by the tapestry at the eye.
Then
is the angle of elevation to the top of the tapestry.
If the observer is standing a distance from the wall on which the
tapestry is hanging, then
and
, from which we find
. The maximum of
occurs at
feet.
- 25.
- A rod that touches the inside corner of the turn and both outside
walls has length
, where is the length
of the portion of the rod from the corner to the wall. The minimum such length
is the length of the longest rod that will go around the turn without
bending. The minimum is . (Note that letting gives the
same solution.)
- 26.
- Minimize
. The minimum
occurs at
.
- 27.
- Minimize the distance function
, replacing
by
. The critical points satisfy the
equation
; the solution
corresponds to
and gives the minimum distance
- 28.
- Minimize
to get
, with the length of the block. That is, the dimmest point
is about 65% of the length of the block from the brighter light.
- 30.
- Let the ladder of length rest against the wall at height
and let its foot be distance from the fence. If the ladder just clears
the fence, then
, which we can use to express the
length of the ladder,
, in terms of just one
variable. Using
, we get
. The minimum occurs at
, and
is
10.81 ft.
- 32.
- (a) Let the ranches be at points and , with point on
the river closest to and point on the river closest to . We are
given , , and , from which it follows that . If we
let the pumping station be at distance from , then the total length of
the pipelines is
. This function has
minimum at
.
(b) The pipeline would run from to to .

(c) Let be the reflection of in the river; then the shortest pipeline would have the same length as the straight line from to , which meets the river at point between and . We can now use similar triangles to find , which is the same as in (a).