Question

Find the maximum and minimum dis-

tances from the origin to the curve

3x2 + 4xy + 6y2 = 140.​

Answer 1

The function achiesves minimum of √20 for (x,y)=(2,4) and (–2,–4) and  achieves a maximum of √70 for (x,y)= (2√14,–√14) and (–2√14,√14).

Given:3x2 + 4xy + 6y2 = 140.​

To prove: the maximum and minimum distances from the origin

solution: The distance of a point (x,y) from the origin on the given curve is √(x²+y²).

We have to maximise and minimise the following function (x²+y²) with the given condition 3x²+4xy+6y²-140=0

F(x,y)= x²+y²+ λ(3x²+4xy+6y²-140)         [Using lagrange function]

Hence,

δF(x,y)/δx =2x+λ(6x+4y)=0

Or, 2x+6λx+4λy=0…(i)

and δF(x,y)/δy=2y+ λ(4x+12y)

Or, 2y+4λx+12λy=0…(ii)

Multiplying above equations we get,

–6λxy+4λy²–4λx²=0

Or, –2λ(3xy–2y²+2x²)=0

Or, (3xy–2y²+2x²)=0

Or, (x+2y)(2x-y)=0

Here, x=–2y…(iii) and x=y/2…(iv)

By substituting values,

x=2√14 when y=–√14

x=–2√14 when y=√14

So, the distance of (2√14,–√14) from the origin

=√70 units.

by substituting values.

y=+4,-4

putting value of y, we get,

x=,-2,2

Moreover ,the distance of (2,4) from the origin, the same with (–2,–4) is √(2²+4²)=√20 units.

the function achiesves minimum of √20 for (x,y)=(2,4) and (–2,–4) and  achieves a maximum of √70 for (x,y)= (2√14,–√14) and (–2√14,√14).

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