The flag should be 192 because 48 x 4 =192
We are given the equations:
<span>5 x + 11 y = 49 -->
eqtn 1</span>
<span>u = 3 x^2 + 6 y^2 -->
eqtn 2</span>
Rewrite eqtn 1 explicit to y:
11 y = 49 – 5 x
<span>y = (49 – 5x) / 11 -->
eqtn 3</span>
Substitute eqtn 3 to eqtn 2:
u = 3 x^2 + 6 [(49 – 5x) / 11]^2
u = 3 x^2 + 6 [(2401 – 490 x + 25 x^2) / 121]
u = 3 x^2 + 14406/121 – 2940x/121 + 150x^2/121
u = 4.24 x^2 – 24.3 x + 119.06
Derive then set du/dx = 0 to get the maxima:
du/dx = 8.48 x – 24.3 = 0
solving for x:
8.48 x = 24.3
x = 2.87
so y is:
y = (49 – 5x) / 11 = (49 – 5*2.87) / 11
y = 3.15
Answer:
<span>George will choose some of each commodity but more y than
x.</span>
First of all, I'm going to assume that we have a concave down parabola, because the stream of water is subjected to gravity.
If we need the vertex to be at 
, the equation will contain a 
term.
If we start with 
we have a parabola, concave down, with vertex at and a maximum of 0.
So, if we add 7, we will translate the function vertically up 7 units, so that the new maximum will be 
We have
Now we only have to fix the fact that this parabola doesn't land at 
, because our parabola is too "narrow". We can work on that by multiplying the squared parenthesis by a certain coefficient: we want
such that:
Plugging these values gets us
As you can see in the attached figure, the parabola we get satisfies all the requests.
Answer: You're not important basically
Step-by-step explanation:
, we have
