# Math Help - Help Needed: Applied Optimization

1. ## Help Needed: Applied Optimization

1. A car rental agency rents 180 cars per day at a rate of 32 dollars per day. For each 1 dollar increase in the daily rate, 5 fewer cars are rented. At what rate should the cars be rented to produce the maximum income, and what is the maximum income?

2. A box is to be made out of a 10 by 16 piece of cardboard. Squares of equal size will be cut out of each corner, and then the ends and sides will be folded up to form a box with an open top. Find the length L, width W, and height H of the resulting box that maximizes the volume. (Assume that WL).

3. A parcel delivery service will deliver a package only if the length plus the girth (distance around) does not exceed 108 inches. Find the maximum volume of a rectangular box with square ends that satisfies the delivery company's requirements.

4. A fence is to be built to enclose a rectangular area of 270 square feet. The fence along three sides is to be made of material that costs 4 dollars per foot, and the material for the fourth side costs 16 dollars per foot. Find the length L and width W (with WL) of the enclosure that is most economical to construct.

5. A small resort is situated on an island. The closest point on the mainland is a point P which is exactly 3 miles from the resort. The closest source of fresh water is 10 miles down the shoreline from P. The resort is planning to lay pipe to bring water from that source. It will run along the shore for some distance and then turn and cross the water to the resort. However, running pipe underwater costs 2.4 times as much as running pipe down the shore. At what distance down the shore from P should the pipe turn and head for the resort?

6.The manager of a large apartment complex knows from experience that 90 units will be occupied if the rent is 370 dollars per month. A market survey suggests that, on the average, one additional unit will remain vacant for each 5 dollar increase in rent. Similarly, one additional unit will be occupied for each 5 dollar decrease in rent. What rent should the manager charge to maximize revenue?

I know this is a lot of problems but my professor hasn't really explained how to solve these effectively and my textbook's examples aren't similar enough to follow. I just have problems setting them up but after that I am usually fine. Thanks for the help..this is due at 6:30PM this evening.

2. Originally Posted by lmao

2. A box is to be made out of a 10 by 16 piece of cardboard. Squares of equal size will be cut out of each corner, and then the ends and sides will be folded up to form a box with an open top. Find the length L, width W, and height H of the resulting box that maximizes the volume. (Assume that WL).
i'm kind of in a rush so i cant help you with everything. here's how to do 2. i assume this is for a calculus class, so i expect you to be able to differentiate something

first draw a diagram. a 16 by 10 rectangle with squares cut out at each corner. label the sides of all the squares x.

you will notice that from each side, we actually cut away 2x (one at each corner).

so the length becomes 16 - 2x
the width becomes 10 - 2x
the height becomes x when you fold it up.

so we have:
L = 16 - 2x
W = 10 - 2x
H = x

V = LWH, where V is the volume.
=> V = (10 - 2x)(16 - 2x)x
=> V = 4x^3 - 52x^2 + 160x

now V' = 12x^2 - 104x + 160

For max, we set V' = 0
=> 12x^2 - 104x + 160 = 0
=> 3x^2 - 26x + 40 = 0

by the quadratic formula, we find that
x = 2 or x = 20/3 ..............now we need to find which of these gives a max. We can do that by finding V''

V'' = 24x - 104 ..............now we plug in both points above
V''(2) = 48 - 104 <0 .............since this is less than zero, we have a concave down and thus we have a maximum. this is an x^3 graph, so we're sure the other critical point will be a minimum.

So for max V, x = 2

therefore H = x = 2
L = 16 - 2x = 16 - 4 = 12
and W = 10 - 2x = 10 - 4 = 6
for maximum volume

3. Hello, lmao!

1. A car rental agency rents 180 cars per day at a rate of 32 dollars per day.
For each 1 dollar increase in the daily rate, 5 fewer cars are rented.
At what rate should the cars be rented to produce the maximum income,
and what is the maximum income?

Let x = number of one-dollar increases in the daily rate.
. . It will charge 32 + x dollars per day.
As a result, it will rent only 180 - 5x cars.

The total income will be: .I .= .(32 + x)(180 - 5x) .= .5760 + 20x - 5x²

To maximize/minimize Income, equate its derivative to zero and solve.
. . I' .= .20 - 10x .= .0 . . x = 2

Hence, they should charge: $32 + 2 .= .$34 per day.
They will rent only: 180 - 10 .= .170 cars
But their daily income is maximized at: 34 x 170 .= .\$5780

(Apply this technique to problem #6.)

4. Hello again, lmao!

Here's #5 . . .

5. A small resort is situated on an island.
The closest point on the mainland is a point P which is exactly 3 miles from the resort.
The closest source of fresh water is 10 miles down the shoreline from P.
The resort is planning to lay pipe to bring water from that source.
It will run along the shore for some distance and then turn and cross the water to the resort.
However, running pipe underwater costs 2.4 times as much as running pipe down the shore.
At what distance down the shore from P should the pipe turn and head for the resort?
Code:
    R *
| *
|   *    ____
3 |     *-√x²+9
|       *
|         *
* - - - - - * - - - - - - - *
P     x     Q     10-x      W

The resort is at R, 3 miles from P on the shore.
The water source is at W: PW = 10.

They will run the pipe from P to point Q on the shore,
. . where: PQ = x, and: QW = 10 - x

Suppose it costs k dollars/mile to lay pipe on land.
. . The cost of the land pipe is: .k(10 - x) dollars.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . _____
From right triangle RPQ, we have: RQ .= .√x² + 9 miles.
At 2.4k dollars/mile to lay underwater pipe,
. . the cost is: .2.4k√x² + 9 dollars.

Hence, the total cost is: .C .= .2.4k(x² + 9)^½ + 10k - kx

Can you finish it now?