Answer:
1250 m²
Step-by-step explanation:
Let x and y denote the sides of the rectangular research plot.
Thus, area is;
A = xy
Now, we are told that end of the plot already has an erected wall. This means we are left with 3 sides to work with.
Thus, if y is the erected wall, and we are using 100m wire for the remaining sides, it means;
2x + y = 100
Thus, y = 100 - 2x
Since A = xy
We have; A = x(100 - 2x)
A = 100x - 2x²
At maximum area, dA/dx = 0.thus;
dA/dx = 100 - 4x
-4x + 100 = 0
4x = 100
x = 100/4
x = 25
Let's confirm if it is maximum from d²A/dx²
d²A/dx² = -4. This is less than 0 and thus it's maximum.
Let's plug in 25 for x in the area equation;
A_max = 25(100 - 2(25))
A_max = 1250 m²
Answer:
85%
Step-by-step explanation:
9+51 = 60
(51/3) ÷ (60/3) = 17/20
(17/20)% = 85%
Answer:
lets start with the easy equation. X+Y=60. so we take 60 and subtract Y to get X by itself. meaning X = 60-Y. every time we see X replace with (60-Y).
X + 5Y = 100 change to
60-Y+5Y = 100
5Y - Y = 4y.
60+4Y = 100. 100 minus the 60 = 40. 4Y=40. divide by 4. Y = 10. ok so now we solve for X. X + Y (or 10) = 60. 60-10 = 50. X = 50.
Step-by-step explanation:
