Calc problem, finding dimensions

[mriley528]

A dairy farmer plans to enclose a rectangular pasture adjacent to a river.
The pasture must contain 180,000 suare meters. No fencing is required along the river. What dimensions will use the smallest amount of fencing?

A= L X W
so 180,000m^2= L X 2W


2. After t years, the value of a car purchased for $20,000 is:

V=20,000(.75)^t

a) What is the value of this car in two years? I put $11,250

B) when will this car be worth $500? I put year 13

c)Use the derivative rule for y=b^x and find the derivative(rate of change) of the value of the car

1) t=1 year

2) t=4 years


Thank you for any help!

 

[tkhunny]

A dairy farmer plans to enclose a rectangular pasture adjacent to a river. The pasture must contain 180,000 suare meters. No fencing is required along the river. What dimensions will use the smallest amount of fencing?

A= L X W
so 180,000m^2= L X 2W

Please define your variables carefully.

L = Pasture Length
W = Pasture Width
L*W = Pasture Area = 180000 m^2
L + 2*W = Required Fencing

Use the Area information to find L in terms of W (or W in terms of L), substitute into the perimeter expression, and finds its minimum value.

 

[tkhunny]

2. After t years, the value of a car purchased for $20,000 is:

V=20,000(.75)^t

a) What is the value of this car in two years? I put $11,250

B) when will this car be worth $500? I put year 13

20000*0.75^2 = 20000*0.5625 = 11250

What are you questioning? If you write it out so YOU can follow, there is nothing to question.

20000*(0.75)^t = 500
0.75^t = 500/20000 = 0.025
log(0.75^t) = log(0.025)
t*log(0.75) = log(0.025)
t = log(0.025)/log(0.75) = 12.82276446

Again, write stuff down so YOU can follow it.

Is "year 13" a sufficient response? You may need "12 years 300 days 7 hours 25 minutes" or something a little less precise. Unfortunately, you have not provided sufficient instruction to decide what kind of answer would be most appropriate.

 

[mriley528]

tkhunny,

for part C, would i use t*log(0.75) = log(0.025), to plug in t=1, and t=4, to find the rate of change?

 

[TchrWill]

A dairy farmer plans to enclose a rectangular pasture adjacent to a river.
The pasture must contain 180,000 suare meters. No fencing is required along the river. What dimensions will use the smallest amount of fencing?

Considering all rectangles with the same perimeter, the square encloses the greatest area.
Proof: Consider a square of dimensions x by x, the area of which is x^2. Adjusting the dimensions by adding a to one side and subtracting a from the other side results in an area of (x + a)(x - a) = x^2 - a^2. Thus, however small the dimension "a" is, the area of the modified rectangle is always less than the square of area x^2.

Using one corner as a center, unfold a 300mx300m square into a rectangle of LxW where L = 600 and W = 300 and you have the smallest amount of fenciing to enclose 180,000 sq.m..

 

[tkhunny]

tkhunny,

for part C, would i use t*log(0.75) = log(0.025), to plug in t=1, and t=4, to find the rate of change?

I don't see a derivative in there, anywhere. Use the rule that is suggested.