A gardener wants to fence the largest possible rectangular area using 200 yards of fencing. Find the best length and width of the garden.
9 answers
please help me!
The largest area is a square.
P = 4S
200/4 = S
P = 4S
200/4 = S
even if it said rectangular?
so its's 50 yd by 50 yd?
so its's 50 yd by 50 yd?
I also need help on this question:
Sandra walked to the top of a hill at a speed of 2km/h, turned around and walked down the hill at a rate of 4km/h. The whole trip took 6 hoours. How many kilometers is it to the top of the hill?
I thought it might be 6km but my math teacher checked it and said it wasn't.
Sandra walked to the top of a hill at a speed of 2km/h, turned around and walked down the hill at a rate of 4km/h. The whole trip took 6 hoours. How many kilometers is it to the top of the hill?
I thought it might be 6km but my math teacher checked it and said it wasn't.
hello?Ms. Sue????
A square is a rectangle. Yes. The largest area would be a square measuring 50 yards on each side.
Her average speed is 3 km/hour.
(3 * 6)/2 = ?
Her average speed is 3 km/hour.
(3 * 6)/2 = ?
W=Width
L=Length
P=Perimeter
A=Area
P=2(W+L)
200=2(W+L) Divide with 2
100=W+L
100-L=W
W=100-L
A=W*L=(100-L)*L
A=100L-L^2
dA/dL=100-2L
Function has minimum or maximum when first derivate=0
100-2L=0
100=2L Divide with 2
50=L
L=50 yd
If second derivate<0 then function has a maximum.
If second derivate>0 then function has a minimum.
Second derivate=-2<0
function has a maximum for L=50 yd
W=100-L
W=100-50=50 yd
A(max)=50*50=2500 yd^2
L=Length
P=Perimeter
A=Area
P=2(W+L)
200=2(W+L) Divide with 2
100=W+L
100-L=W
W=100-L
A=W*L=(100-L)*L
A=100L-L^2
dA/dL=100-2L
Function has minimum or maximum when first derivate=0
100-2L=0
100=2L Divide with 2
50=L
L=50 yd
If second derivate<0 then function has a maximum.
If second derivate>0 then function has a minimum.
Second derivate=-2<0
function has a maximum for L=50 yd
W=100-L
W=100-50=50 yd
A(max)=50*50=2500 yd^2
what do you do with (*)
* is sign for multiply