Answer:
All I can really do is state it differently <> 4.5q /9
Answer:
Step-by-step explanation:
Question is not clear.
Base is 5 units
height is 3 units
area is 7.5 units²
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!
Step-by-step explanation:
9x²+bx+ 100
(3x)²+ bx + 10²
(3x)² + 10² + bx
(3x)² + 10² + bx