x^2 y = V
G = 4x+y = 6
==================
y = 6-4x
x^2 (6-4x) = V
now do the complete the square thing like I did in your other problem to find the vertex of the parabola
The size of a parcel despatched through the post used to be limited by the fact that the sum of its length and girth (perimeter of the cross section) must not exceed 6 feet. What was the volume of the largest parcel of square cross- section which was acceptable for posting? (Let the cross-section be a square of side x feet.)
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