Basically, what this asks you is to maximize the are A=ab where a and b are the sides of the recatangular area (b is the long side opposite to the river, a is the short side that also is the common fence of both corrals). Your maximization is constrained by the length of the fence, so you have to maximize subject to 3a+b=450 (drawing a sketch helps - again, b is the longer side opposite to the river, a are the three smaller parts restricting the corrals)
3a+b = 450
b = 450 - 3a
so the maximization max(ab) becomes
max(a(450-3a)=max(450a-3a^2)
Since this is in one variable, we can just take the derivative and set it equal to zero:
450-6a=0
6a=450
a=75
Plugging back into b=450-3a yields
b=450-3*75
b=450-225
b=215
Hope that helps!
Help what ??? I don’t see anything
Answer:
buy the cheapest one with more band aids.
Step-by-step explanation:
Answer:
Step-by-step explanation:
Begin
If this is a rhombus then <8 = 90 degrees as do all the central angles. That's because the diagonals intersect at right angles.
<4 = 38 z formation for parallel lines.
<7 = 52 The angles are part of a right angle triangle. <7 +38 = 90
<2 = 52 z formation of parallel lines (a rhombus has ll lines).
<3 = 38 The diagonals of a rhombus bisect each other.
<5 = 38 z formation for parallel lines.
<6 = 52 The diagonals of a rhombus are angle bisectors.
<1 = 52 The diagonals of a rhombus are angle bisectors.
