how could you find the measurements of the frame with a 16 inch perimeter antd the greatest possible area?
2 answers
A square of 4 by 4 would give you the largest area
P=Perimeter
A=Area
a,b=sides of rectacangle
P=2a+2b
P=2(a+b) Divide with 2
P/2=a+b
a+b=P/2
a+b=16/2
a+b=8
b=8-a
A=a*b
A=a*(8-a)
A=8a-a^2
A=-a^2+8a
Function have extreme value in point where first derivative=0
First derivative:
dA/da=-2a+8
dA/da=0
-2a+8=0
8=2a Divide with 2
a=4in
b=8-a
b=8-4
b=4in
Second derivative:
d^2A/da^2=-2
When second derivative is negative fuction have maximum.
a=4in b=4 in
A(max)=4*4
A(max)=16in^2
A=Area
a,b=sides of rectacangle
P=2a+2b
P=2(a+b) Divide with 2
P/2=a+b
a+b=P/2
a+b=16/2
a+b=8
b=8-a
A=a*b
A=a*(8-a)
A=8a-a^2
A=-a^2+8a
Function have extreme value in point where first derivative=0
First derivative:
dA/da=-2a+8
dA/da=0
-2a+8=0
8=2a Divide with 2
a=4in
b=8-a
b=8-4
b=4in
Second derivative:
d^2A/da^2=-2
When second derivative is negative fuction have maximum.
a=4in b=4 in
A(max)=4*4
A(max)=16in^2
5 answers