Mathematics: How many individual peanuts, unshelled (not counted in onces or cups or anythign) would fit in a 14 oz cup?

How many individual peanuts, unshelled (not counted

hetdashadia123

27.02.2021

Considering the spaces in the coefficients, I'm guessing the plane has equation

$x+\dfrac13y+\dfrac18z=1$

Let (x,y,z) be the vertex of the box that lies in this plane. The box has volume

V(x,y,z)=xyz

and we want to maximize this subject to the plane equation. Using the method of Lagrange multipliers, the Lagrangian is

$L(x,y,z,\lambda)=xyz+\lambda\left(x+\dfrac13y+\dfrac18z-1\right)$

with partial derivatives (set equal to 0)

$L_x=yz+\lambda=0$

$L_y=xz+\dfrac13\lambda=0$

$L_z=xy+\dfrac18\lambda=0$

$L_\lambda=x+\dfrac13y+\dfrac18z-1=0$

We have

$xL_x+yL_y+zL_z=3xyz+\lambda\left(x+\dfrac13y+\dfrac18z\right)=0$

$L_\lambda=0\implies3xyz+\lambda=0$

$L_x=0\implies3xyz-yz=yz(3x-1)=0\implies x=\dfrac13$

(we ignore y=0 and z=0 because those values correspond to a degenerate box with zero volume)

$L_y=0\implies3xyz-3xz=3xz(y-1)=0\implies y=1$

$L_z=0\implies3xyz-8xy=xy(3z-8)=0\implies z=\dfrac83$

So we get a maximum volume of $V\left(\dfrac13,1,\dfrac83\right)=\dfrac89$ .

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