# Calculus Optimization Word Prob

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• Dec 4th 2006, 05:24 PM
drewms64
Calculus Optimization Word Prob
Hi, I have 2 questions on these words problems. I really am not sure where to start these.

1. An advertisement consists of a rectangular printed region with margins of 2 inches each at the top and bottom and 1 inch at each side. If the area of the printed region is to be 98 in^2 , find the overall dimensions if the total area of the advertisement is to be a minimum.

2. A rancher is going to build a 3-sided enclosure with a divider down the middle. The cost per foot of the 3 side walls are $6/ft, with the single back wall being$10/ft. The area enclosed will be 180 ft^2. What dimensions would minimize the cost?
The pen looks like this.
____
| | |

for this one I had an idea of setting 180ft^2 = 3y+x but am not sure if this is the right way to start.
• Dec 4th 2006, 09:47 PM
CaptainBlack
Quote:

Originally Posted by drewms64
1. An advertisement consists of a rectangular printed region with margins of 2 inches each at the top and bottom and 1 inch at each side. If the area of the printed region is to be 98 in^2 , find the overall dimensions if the total area of the advertisement is to be a minimum.

Let the length and bredth of the ad be $l$ and $b$, then:

$(l-4)(b-2)=98$

rearranging:

$
l=\frac{98}{b-2}+4
$

Then the total area of the ad is:

$
A=l.b=\left(\frac{98}{b-2}+4 \right).b
$

So now you just need to find the $b$ which minimises $A$

RonL
• Dec 4th 2006, 09:51 PM
CaptainBlack
Quote:

Originally Posted by drewms64
2. A rancher is going to build a 3-sided enclosure with a divider down the middle. The cost per foot of the 3 side walls are $6/ft, with the single back wall being$10/ft. The area enclosed will be 180 ft^2. What dimensions would minimize the cost?
The pen looks like this.
____
| | |

for this one I had an idea of setting 180ft^2 = 3y+x but am not sure if this is the right way to start.

(all dimensions in ft and areas in sq ft)

The area:

$
A=s \times b=180
$

where $s$ is the length of the side walls, and $b$ is the length of the back wall.

The cost is:

$
C=b \times 10+ 3 \times s \times b
$

So you now eliminate one of $b$ or $s$ from the expression for $C$ using the area constraint, then minimise $C$.

RonL
• Dec 5th 2006, 05:23 PM
drewms64
Quote:

Originally Posted by CaptainBlack
Let the length and bredth of the ad be $l$ and $b$, then:

$(l-4)(b-2)=98$

rearranging:

$
l=\frac{98}{b-2}+4
$

Then the total area of the ad is:

$
A=l.b=\left(\frac{98}{b-2}+4 \right).
$

So now you just need to find the $b$ which minimises $A$

RonL

Thanks for the advice on both of the questions, for this one would it then end up being,
x=b y=l
$
A(y)= \left(\frac{98}{y-4}+2 \right)y
$

$
= \left(\frac{98y}{y-4}+2y \right)
$

$
A'(y)= \left(\frac{(y-4)(98)-(98y)(1)}{(y-4)^2)}+2 \right)
$

$
= \left(\frac{-392}{(y-4)^2}+2 \right) = 0
$

not sure how to get [tex] to work for this one, but

-392(y-4)^-2 = -2
(y-4)^-2 = 196
take the -2 root of each side
y-4 = - 14
y = -10

then when I find x I get -5

I dont think these should be negative though?
• Dec 5th 2006, 06:05 PM
alinailiescu
Pay attention to square root
Quote:

Originally Posted by drewms64
Thanks for the advice on both of the questions, for this one would it then end up being,
x=b y=l
$
A(y)= \left(\frac{98}{y-4}+2 \right)y
$

$
= \left(\frac{98y}{y-4}+2y \right)
$

$
A'(y)= \left(\frac{(y-4)(98)-(98y)(1)}{(y-4)^2)}+2 \right)
$

$
= \left(\frac{-392}{(y-4)^2}+2 \right) = 0
$

not sure how to get [tex] to work for this one, but

-392(y-4)^-2 = -2
(y-4)^-2 = 196
take the -2 root of each side
y-4 = - 14
y = -10

then when I find x I get -5

I dont think these should be negative though?

There is not such thing as -2 root. You should apply the cross product rule:a/b=c/d is equivalent to a*d=b*c.
therefore, (y-4)^-2=196 becomes (y-4)^2=1/196 and now apply the square root to find
y-4=1/14 or -1/14
So, your solutions are y=4+1/14 and y=4-1/14, both positive.
• Dec 5th 2006, 06:58 PM
drewms64
Quote:

Originally Posted by alinailiescu
There is not such thing as -2 root. You should apply the cross product rule:a/b=c/d is equivalent to a*d=b*c.
therefore, (y-4)^-2=196 becomes (y-4)^2=1/196 and now apply the square root to find
y-4=1/14 or -1/14
So, your solutions are y=4+1/14 and y=4-1/14, both positive.

By doing this I got y= 57/14 and by plugging that into the x = (98/(y-4))+2 I get 1372. That seems wrong for the area being 98in^2
• Dec 6th 2006, 07:14 AM
alinailiescu
Correction
Quote:

Originally Posted by drewms64
By doing this I got y= 57/14 and by plugging that into the x = (98/(y-4))+2 I get 1372. That seems wrong for the area being 98in^2

Sorry I did not read all of your solution.Your mistake was a little bit sooner.
-392(y-4)^-2=-2
(y-4)^-2=-2/-392
(y-4)^-2=1/196
(y-4)^2=196
or y-4=14 or -14
so, y=18
x=9
The dimmension of the printed area being 14, respectively 7 give as the required 98 in^2