d = [(x-5)² + (y-5)² + (z - 0)²]^1/2

d(x,y) = [(x-5)² + (y-5)² + (x²+y²)²]^1/2

d² = f(x,y) = (x-5)² + (y-5)² + (x²+y²)²

     Solve the system below.

f_x = 2(x-5) + 2(x²+y²)(2x) = 0

f_y = 2(y-5) + 2(x²+y²)(2y) = 0

     y(x-5) + y(x²+y²)(2x) = 0

     x(y-5) + x(x²+y²)(2y) = 0

Subtraction produces -5y + 5x = 0 which yields

     x = y

Back substitution produces

     x - 5 + 2x(2x²) = 0

     4x³ + x - 5 = 0

The solutions are 

x = 1, x = -1/2 - i and x = -1/2 + 1

For x=1, y=1, and z=2.

The point on the graph of z = x² + y² closest to the point (5,5,0) is (1,1,2).  d(1,1) = 6 

The graph at the right is the graph of d(x,y).  Click on the graph to see it rotate.

Let x = width of base, y = length of base, z = height

3xy + 4xz + 4yz = 100

z = (100 - 3xy)/(4x + 4y)     (1)

V = xyz

V(x,y) = xy(100 - 3xy)/(4x + 4y)

V_x = (400y² - 12x²y² - 24xy³)/(4x + 4y)² = 0

V_y = (400x² - 12x²y² - 24yx³)/(4x + 4y)² = 0

This leads to the system

100 - 3x² - 6xy = 0

100 - 3y² - 6xy = 0

and subtraction leads to x = y.

Back substitution produces

100 - 9x² = 0 so x = 10/3 and y = 10/3

Substituting into (1) above produces z = 5/2.

Thus the volume is a maximum if the dimensions are

10/3 ft by 10/3 ft by 5/2 ft 

and the maximum volume is 250/9 cubic ft.

The graph at the right is the graph of V(x,y).  Click on the graph to see it rotate.