Answer:
The required width of the field that would maximize the area is = 1250 feet
Step-by-step explanation:
Given that:
The total fencing length = 5000 ft
Let consider w to be the width and L to be the length.
Then; the perimeter of the rectangular field by assuming a parallel direction is:
P = 3L + 2w
⇒ 3L + 2w = 5000
3L = 5000 - 2w
![L = \dfrac{5000}{3} - \dfrac{2w}{3}](https://tex.z-dn.net/?f=L%20%3D%20%5Cdfrac%7B5000%7D%7B3%7D%20-%20%5Cdfrac%7B2w%7D%7B3%7D)
Recall that:
The area of the rectangle = L×w
![A(w) = ( \dfrac{5000}{3}-\dfrac{2}{3} ) w](https://tex.z-dn.net/?f=A%28w%29%20%3D%20%28%20%5Cdfrac%7B5000%7D%7B3%7D-%5Cdfrac%7B2%7D%7B3%7D%20%29%20w)
![A(w) = \dfrac{5000}{3}w-\dfrac{2}{3} w^2](https://tex.z-dn.net/?f=A%28w%29%20%3D%20%20%5Cdfrac%7B5000%7D%7B3%7Dw-%5Cdfrac%7B2%7D%7B3%7D%20%20w%5E2)
Taking the differentiation of both sides with respect to t; we have:
![A' (w) = \dfrac{5000}{3} - \dfrac{2}{3} ( 2 w)](https://tex.z-dn.net/?f=A%27%20%28w%29%20%3D%20%5Cdfrac%7B5000%7D%7B3%7D%20-%20%5Cdfrac%7B2%7D%7B3%7D%20%28%202%20w%29)
![A' (w) = \dfrac{5000}{3} - \dfrac{4w}{3}](https://tex.z-dn.net/?f=A%27%20%28w%29%20%3D%20%5Cdfrac%7B5000%7D%7B3%7D%20-%20%5Cdfrac%7B4w%7D%7B3%7D)
Then; we set A'(w) to be equal to zero;
So; ![\dfrac{5000}{3} - \dfrac{4w}{3}=0](https://tex.z-dn.net/?f=%5Cdfrac%7B5000%7D%7B3%7D%20-%20%5Cdfrac%7B4w%7D%7B3%7D%3D0)
5000 = 4w
w = 5000/4
w = 1250
Thus; the required width of the field that would maximize the area is = 1250 feet
Also, the length
can now be :
![L = \dfrac{5000}{3} - \dfrac{2(1250)}{3}](https://tex.z-dn.net/?f=L%20%3D%20%5Cdfrac%7B5000%7D%7B3%7D%20-%20%5Cdfrac%7B2%281250%29%7D%7B3%7D)
L = (5000 -2500)/3
L = 2500/3 feet
Suppose, the farmer divides the plot parallel to the width; Then 2500/3 feet = 833.33 feet and the length L = 1250 feet.
y = 5 - 7
=> y = -2 -------(i)
y = -2x + 7
=> -2 = -2x + 7 [From (i)]
=> -2 -7 = -2x
=> -9 = -2x
=> -9/-2 = x
=> 9/2 = x
So, x = 9/2 and y = -2
Answer: 92.67
Step-by-step explanation:
In the Screen Shot is the answer
Answer:
Domain: 0
t
40 Range: 0
V(t)
200
Step-by-step explanation:
The domain is greater than or equal to 0 and less than or equal to 40 because in 40 months, she will lose all her money ($5 x 40months = $200) since she loses 5 dollars per month. Not to mention, time can't be negative in any situation. For the range, the value has to be greater than or equal to zero and less than or equal to 200 because she only has $200, the more time she spends using the phone, the more money she will lose over the course of the month( limiting her time). I hope this helps!