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Two contour lines can intersect. A perfect example is a Lagrange Multiplier which is encountered in Calculus III. We are given a function that has restraints (side conditions). An optimization engineer working for a box factory might be asked to find the maximum volume of a cardboard box given the restraint that it has a surface area of 1500 cm2 and a total edge length of 200 cm.

We are seeking the extreme values of f(x,y,z) that lie on the one of the level curves (c) of g(x,y,z) and h(x,y,z). These occur at a point P(x0,y0,z0) where you can find the highest level surfaces (k) of f(x,y,z) that are intersected by the level curves (c) of g(x,y,z) and h(x,y,z). These intersections occur when they just barely touch one another. Meaning they have a common tangent line. Further, their normal lines are the same, implying that their gradient vectors ∇f, ∇g, ∇h are parallel.

f = λ∇g + μ∇h. This works if ∇g and ∇h0.

Eq. 1 f: V=xyz

Eq. 2 g: 1500=2(xy)2+2(xz)2+2(yz)2

Eq. 3 h: 200=√x2+y2+z2

f =(yz,xz,xy)

g = (4xy2+4xz2,4x2y+4yz2,4x2z+4y2z)

h = (x/√x2+y2+z2, y/√x2+y2+z2, z/√x2+y2+z2)

Eq. 4 yz= λ(4xy2+4xz2) + μ(x/√x2+y2+z2)

Eq. 5 xz= λ(4x2y+4yz2) + μ(y/√x2+y2+z2)

Eq. 6 xy= λ(4x2z+4y2z) + μ(z/√x2+y2+z2)

We have 6 equations and 6 unknowns (x,y,z,λ,μ and V). We will have to use back substitution to solve.

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