Thread: 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.

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 and , then:



rearranging:



Then the total area of the ad is:



So now you just need to find the which minimises

RonL

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:



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

The cost is:



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

RonL

Originally Posted by CaptainBlack

Let the length and bredth of the ad be and , then:



rearranging:



Then the total area of the ad is:



So now you just need to find the which minimises

RonL

Thanks for the advice on both of the questions, for this one would it then end up being,
x=b y=l








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?

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








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.

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

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