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:
Four billion, two-hundred seventy five thousand.
Find and graph the feasible region for the following constraints: x + y < 5. 2x<span> + y > 4 ... y = 10/3. x = 30/3 - 10/3 = 20/3. Intersects at (20/3, 10/3). -x + </span>2y<span> = 0. x - </span>2y = 0.
435 being the constant means that is what they started with
Hope this helps :)
