For this case, the area is given by:
A = x * (200-2x)
Rewriting:
A = 200x-2x ^ 2
Deriving the expression we have:
A '= 200-4x
Equaling zero we have:
200-4x = 0
We clear x:
4x = 200
x = 200/4
x = 50 feet
Then, the maximum area is:
A (50) = 50 * (200-2 * 50)
A (50) = 5000 feet ^ 2
Answer:
the maximum possible area that can be enclosed with his 200ft of fencing is:
A (50) = 5000 feet ^ 2
Answer:
Step-by-step explanation:
<u>Given Data:</u>
Base area = 
= 108 in.²
Volume = V = 729 in.³
<u>Required:</u>
Height = h = ?
<u>Formula:</u>
<u>Solution:</u>
For h, rearranging formula:
![\displaystyle h = \frac{V}{A_{B}} \\\\h =\frac{729 \ in.^3}{108 \ in.^2} \\\\h = 6.75 \ in.\\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20h%20%3D%20%5Cfrac%7BV%7D%7BA_%7BB%7D%7D%20%5C%5C%5C%5Ch%20%3D%5Cfrac%7B729%20%5C%20in.%5E3%7D%7B108%20%5C%20in.%5E2%7D%20%5C%5C%5C%5Ch%20%3D%206.75%20%5C%20in.%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
In general if multiplying, move decimal to RIGHT by whatever the exponent is.
In this example 10^1, the exponent = 1
9.4 with decimal moved 1 place to right = 94
54% = 54/100 = 27/50
Answer: 27/50
Hope it helped :)
