Expert: Scotto - 4/28/2009

Question
1. Find the dimensions of the largest rectangle whose perimeter is 3200 feet. (Enter the dimensions from smallest to largest.)
Side:
Side:
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2. A farmer wishes to build a large rectangular grazing area of dimensions, width, w, and length, l,splitting the area in to 4 equal fields by dividing the width into 4 equal pieces. The total area of all the fields must be 10000 square feet. Find the dimensions that uses the least amount of fencing.
(It is not always the case that the width is shorter than the length. Round-off your answers to 3 decimal places.)

Width w =
Length l =
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3. A large bin for holding heavy material must be in the shape of a box with an open top and a square base. The base will cost 6 dollars a square foot and the sides will cost 8 dollars a foot. If the volume must be 110 cubic feet. Find the dimensions that will minimize the cost of the box's construction.

Base :
Each side :
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4. A 30 by 30 square piece of cardboard is to be made into a box by cutting out equal square corners from each side of the square. What size corners should be cut out so that the volume of the box is maximized?

Length of each corner :
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5. A rectangular page is to contain 155 square inches of print. The top and bottom margins are each 0.8 inches wide , and the margins on each side is 1.5 inches wide. What should the dimensions be if the least amount of material is to be used ?

Length of top(bottom) : __ inches
Length of side : ___ inches
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6. Suppose postal requirements are that the maximum of the length plus the girth (cross sectional perimeter) of a rectangular package that may be sent is 225 inches. Find the dimensions of the package with square ends whose volume is to be maximum.

Square side :
Length :
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7.
Bob is at point B 30 miles from A. Alice is in a boat in the sea at point C 1.5 miles from the beach. Alice rows at 2 miles per hour and walks at 4.25 miles per hour , where along the beach should she land so that she may get to Bob in the least amount of time ?

Distance of landing point measured from A : ___ miles.  

Answer
1. Find the dimensions of the largest rectangle whose perimeter is 3200 feet. (Enter the dimensions from smallest to largest.)
Side:
Side:

3200 = LW where L is length and W is width.
The perimeter P = 2L + 2W.
From the first equation, we know that W = 3200/L,
so P = 2L + 6400/L.
Take the derivatve of P, set to 0, and solve for L.  That's one side.
Note that W=3200/L.  That's the other side.

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2. A farmer wishes to build a large rectangular grazing area of dimensions, width, w, and length, l,splitting the area in to 4 equal fields by dividing the width into 4 equal pieces. The total area of all the fields must be 10000 square feet. Find the dimensions that uses the least amount of fencing.
(It is not always the case that the width is shorter than the length. Round-off your answers to 3 decimal places.)

To do so, he must run the fence on one side, the other side, and down the middle in both directions.  
That makes the fence length Y = 3L + 3W.
We also are given the LW = 10,000.
This tells us the W = 10,000 / L.
Putting this back in a couple of rows earlier into the fence length
gives Y = 3L + 30,000 / L.
Find Y', set to 0, and solve for L.
Once L is known, W = 10,000 / L, so W can be found as well.

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3. A large bin for holding heavy material must be in the shape of a box with an open top and a square base. The base will cost 6 dollars a square foot and the sides will cost 8 dollars a foot. If the volume must be 110 cubic feet. Find the dimensions that will minimize the cost of the box's construction.

The volume must be 110 cubic feet.  Let H = height and W = width.
The volume V is W²H, since the base is square.
The bottom area is W².  The sides ( 4 of them ) have area 4WL.
The cost is then C = 6W² + 8WH.

From the volume V, which is 110, we know that 110 = W²H.
This says that H = 110/W².
Putting this into cost gives us C = 6W² + 880/W.
To find W, find dC/dW and set it to 0.
Note that H = 110/W², so H can be found as well.
Remember the base was W² and each side was WH.

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4. A 30 by 30 square piece of cardboard is to be made into a box by cutting out equal square corners from each side of the square. What size corners should be cut out so that the volume of the box is maximized?

Lets say that after we cut a square piece out of each corner,
we have X feet left over.  That would make each corner have length
(30 - X/2).  The volume of this box would be X by X by (30 - X/2).
In other words, V = (30 - X/2)X².
Multiply V out, giving 30X² - X^3/2.
Find dV/dX, set it to 0, solve for X.
The length of each corner, remember, was (30 - X/2).

Note: X was used as a side to make the volume involve a X²,
not a (30-2Y)² where Y is the length of each corner.
Now this requires a little computation at the end,
but either method would work.

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5. A rectangular page is to contain 155 square inches of print. The top and bottom margins are each 0.8 inches wide , and the margins on each side is 1.5 inches wide. What should the dimensions be if the least amount of material is to be used ?

Lets say that the printing takes up and area of width W and height H.
The paper has top and bottom of 0.8, making the height H + 2(0.8) =
H + 1.6.  The paper has a left and right margin of 1.5, making the
width into W + 2(1.5) = W + 3.

The total area is A = (H + 1.6)(W + 3).
This can be multiplied out.

Now we are told that WH = area = 155 square inches.
What this says is that W = 155/H.
Substitute that into the equation for A
( multiply the equation out first seems easier to me).
We now have an equation in terms of H alone.
Find dA/dH, set to 0, and solve for H.
Note that W = 155/H, so we now know W.
W is the width of the page, which is length of the top/bottom.
H is the height of the page, which is the length of the sides.

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6. Suppose postal requirements are that the maximum of the length plus the girth (cross sectional perimeter) of a rectangular package that may be sent is 225 inches. Find the dimensions of the package with square ends whose volume is to be maximum.

Let L be the length, G be the girth.
We know that L + G must be less thatn 225.  
To find the limit for the biggest L and G can be, set L + G = 225.
By cross sectional perimeter, I assume you mean 2*width + 2*height.
Let W be the width and H be the height.

This problem says that it is square, so W=H,
and we have 4W as the perimter.

Out equation for size is then L + 4W = 225 and volume V = LW².

It can be seen that L = 225 - 4W, so put this into the equation for V to get V = (225 - W)W².  Multiply this equation out, find dV/dW,
set it equal to 0, solve for W, solve for L.
The area of a square side is W².  The length is L.

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7.
Bob is at point B 30 miles from A. Alice is in a boat in the sea at point C 1.5 miles from the beach. Alice rows at 2 miles per hour and walks at 4.25 miles per hour , where along the beach should she land so that she may get to Bob in the least amount of time ?

Distance of landing point measured from A : ___ miles.

I will assume that Alice is in a boat that is straight out from A.
The distance out is 1.5 miles.

Make the landing point L.
L is X miles from A and 30-X miles from B.
The distance to A from L is √(X²+1.5²).
The time it takes for Alice to get to B is
T = (30-X)/4.25 + √(X²+1.5²)/2.
Find dT/dX, set to 0.  Note that when finding the derivative of
√(X²+1.5²)/2, it can be thought of as (X²+1.5²)^0.5/2.
The derivative is (0.5*(same thing)^-0.5/2)(derivative of what is beneath the squareroot sign).  That derivative of what is beneath the squareroot sign is 2x.  So the derivatve has 2x*0.5/2 = x/2, all over √(X²+1.5²).

To find the distance to the land,
first find the distance on the land.
This is 30-X.
The distance to Alice is √((30-X)² + 1.5²).
That's √(900 - 2X + X² + 2.25).  First you have to find X.