X=0, y=6
x=1, y=4
x=2, y=2
x=3, y=0
x=4, y=-2
Hope this helps :)
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: what’s your question?
Step-by-step explanation:
Okay, first we need to figure out what one would make. We need to figure out the area of the base x height. The diameter is 8 so we will cut that to 4 so we can do piR2! 3.14x16 would be 50.24. Then we mutiply that by 10 to get 502.4cm3! Okay, now we would divide 2000/502.4 to find out how many times we would have to make the containers to get to 2000. The answer is 3.98, which we will have to round up since its asks for the minimum and you cant make a container 3.98 times. So the answer is 4!
