Two chicken coops are to be built adjacent to one another from 168 ft of fencing. What dimensions should be used to maximize the
area of an individual coop?
1 answer:
Answer:
21 ft by 28 ft
Step-by-step explanation:
To maximize the area, see the attached.
Perimeter will be 4l+3w which is equal to the fencing perimeter, given as 168
4l+3w=168
Making l the subject then
4l=168-3w
l=42-¾w
Area of individual land will be lw and substituting l with l=42-¾w
Then
A=lw=(42-¾w)w=42w-¾w²
A=42w-¾w²
Getting the first derivative of the above with respect to w rhen
42-w6/4=0
w6/4=42
w=42*4/6=28
Since
l=42-¾w=42-¾(28)=21
Therefore, maximum dimensions are 21 for l and 28 for w
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There are 62 quarters and 52 dimes in the jar.
Step-by-step explanation:
Given,
Worth of jar coins = $20.70 = 20.70*100 = 2070 cents
1 Quarter = 25 cents
1 Dime = 10 cents
Let,
Number of quarters = x
Number of dimes = y
According to given statement;
25x+10y=2070 Eqn 1
x=y+10 Eqn 2
Putting value of x from Eqn 2 in Eqn 1
Dividing both sides by 35
Putting y=52 in Eqn 2
There are 62 quarters and 52 dimes in the jar.
Keywords: linear equation, substitution method
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