X to the third power multiple by x to the third power
2 answers:
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X+y=32
xy=-80
x+y=32
minus x both sides
y=32-x
sub for y
x(32-x)=-80
32x-x^2=-80
times both sides by -1
x^2-32x=80
minus 80 both sides
x^2-32x-80=0
use quadratic formula
where
ax^2+bx+c=0
x=
a=1
b=-32
c=-80
x=
x=
x=
x=
x=16+/-4√21
the numbers are
16+4√21 and 16-4√21
xy=-80
x+y=32
minus x both sides
y=32-x
sub for y
x(32-x)=-80
32x-x^2=-80
times both sides by -1
x^2-32x=80
minus 80 both sides
x^2-32x-80=0
use quadratic formula
where
ax^2+bx+c=0
x=
a=1
b=-32
c=-80
x=
x=
x=
x=
x=16+/-4√21
the numbers are
16+4√21 and 16-4√21
Answer:
Length (parallel to the x-axis): ;
Height (parallel to the y-axis): .
Step-by-step explanation:
Let the top-right vertice of this rectangle .
. The opposite vertice will be at
. The length the rectangle will be
while its height will be
.
Function that needs to be maximized: .
The rectangle is inscribed in the ellipse. As a result, all its vertices shall be on the ellipse. In other words, they should satisfy the equation for the ellipse. Hence that equation will be the equation for the constraint on x and y.
For Lagrange's Multipliers to work, the constraint shall be in the form: . In this case
.
Start by finding the first derivatives of and
with respect to x and y, respectively:
,
.
,
.
This method asks for a non-zero constant, , to satisfy the equations:
, and
.
(Note that this method still applies even if there are more than two variables.)
That's two equations for three variables. Don't panic. The constraint itself acts as the third equation of this system:
.
.
Replace the in equation
with the right-hand side of equation
.
.
Before dividing both sides by , make sure whether
.
If , the area of the rectangle will equal to zero. That's likely not a solution.
If , divide both sides by
,
. Hence by equation
,
. Replace the
in equation
with this expression to obtain (given that
)
. Hence
. The length of the rectangle will be
while the height will be
. If there's more than one possible solutions, evaluate the function that needs to be maximized at each point. Choose the point that gives the maximum value.