(2x+3y)⁴
1) let 2x = a and 3y = b
(a+b)⁴ = a⁴ + a³b + a²b² + ab³ + b⁴
Now let's find the coefficient of each factor using Pascal Triangle
0 | 1
1 | 1 1
2 | 1 2 1
3 | 1 3 3 1
4 | 1 4 6 4 1
0,1,2,3,4,.. represent the exponents of binomials
Since our binomial has a 4th exponents, the coefficients are respectively:
(1)a⁴ + (4)a³b + (6)a²b² + (4)ab³ + (1)b⁴
Now replace a and b by their real values in (1):
2⁴x⁴ +(4)8x³(3y) + (6)(2²x²)(3²y²) + (4)(2x)(3³y³) + (1)(3⁴)(y⁴)
16x⁴ + 96x³y + 216x²y² + 216xy³ + 81y⁴
It would be: k+3 = 10
subtracting 3 from both sides,
k+3-3 = 10-3
k=7
Notice that

so the constraint is a set of two lines,

and only the first line passes through the first quadrant.
The distance between any point
in the plane is
, but we know that
and
share the same critical points, so we need only worry about minimizing
. The Lagrangian for this problem is then

with partial derivatives (set equal to 0)



We have

which tells us that

so that
is a critical point. The Hessian for the target function
is

which is positive definite for all
, so the critical point is the site of a minimum. The minimum distance itself (which we don't seem to care about for this problem, but we might as well state it) is
.
Answer:
The first step is to multiply by a power of 10, so the divisor is a whole number.
Step-by-step explanation: