Answer:
Perimeter: P=62 feet
P=2(b+h)
62=2(b+h)
Dividing both sides of the equation by 2:
62/2=2(b+h)/2
31=b+h
b+h=31 (1)
Area: A=bh (2)
Isolating h in equation (1)
(1) b+h=31→b+h-b=31-b→h=31-b (3)
Replacing h by 31-b in equation (2)
(2) A=bh
A=b(31-b)
A=31b-b^2
To maximize the area:
A'=0
A'=(A)'=(31b-b^2)'=(31b)'-(b^2)'=31-2b^(2-1)→A'=31-2b
A'=0→31-2b=0
Solving for b:
31-2b+2b=0+2b
31=2b
Dividing both sides by 2:
31/2=2b/2
31/2=b
b=31/2=15.5
Replacing b by 31/2 in equation (3)
h=31-b
h=31-31/2
h=(2*31-31)/2
h=(62-31)/2
h=31/2
The dimensions are 31/2 ft x 31/2 ft = 15.5 ft x 15.5 ft
The area with these dimensions is: A=(15.5 ft)(15.5 ft)→A=240.25 ft^2
These dimensions are not in the options
1) The first option has an area of: A=(18 ft)(13 ft)→A=234 ft^2
2) The second option has an area of: A=(19 ft)(12 ft)→A=228 ft^2
3) The third option has an area of: A=(17 ft)(14 ft)→A=238 ft^2
The third option has the largest area.
Answer: Third option
Read more on Brainly.com - brainly.com/question/9630010#readmore
