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!
60 inches since there is 2 inches of snow for 3 days, so that equals 6 inches for 3 days and then you just have to multiply 6 by 10 and you get 60 inches of snow, or 5 feet
So if you solve for n
4n^2+3=7n
make into trinomial
subtract 7n from both sides
4n^2-7n+3=0
if we can factor this, then we can asume that the factors are equal to zero becase if
xy=0 then assume x and/or y=0 so
to factor an equation in ax^2+bx+c where a is greater than 1 then
b=t+z
a times c=t times z
to factor 4n^2-7n+3 you do
4 times 3=12
what 2 numbers multiply to get 12 and add to get -7
the numbers are -3 and -4 so
split up the -7
4n^2-4n-3n+3
group
(4n^2-4n)+(-3n+3)
undistribute
(4n)(n-1)+(-3)(n-1)
reverse distributive property
ab+ac=a(b+c)
(4n-3)(n-1)=0
set each to zero
4n-3=0
add 3 to both sides
4n=3
divide both sides by 4
n=3/4
n-1=0
add 1 to both sides
n=1
n=3/4 or 1
