Kenneth Kuttler

488 CHAPTER 23. OPTIMIZATION

Example 23.1.6 Find the volume of the smallest tetrahedron made up of the coordinateplanes in the first octant and a plane which is tangent to the sphere x2 + y2 + z2 = 4.

The normal to the sphere at a point (x0,y0,z0) of the sphere is(x0,y0,

√4− x2

0 − y20

)and so the equation of the tangent plane at this point is

x0 (x− x0)+ y0 (y− y0)+√

4− x20 − y2

(z−√

4− x20 − y2

)= 0

When x = y = 0, z = 4√(4−x2

0−y20)

. When z = 0 = y, x = 4x0

, and when z = x = 0, y = 4y0

Therefore, the function to minimize is

f (x,y) =16

xy√(4− x2 − y2)

This is because in beginning calculus it was shown that the volume of a pyramid is 1/3 thearea of the base times the height. Therefore, you simply need to find the gradient of thisand set it equal to zero. Thus upon taking the partial derivatives, you need to have

−4+2x2 + y2

x2y(−4+ x2 + y2)√(4− x2 − y2)

= 0,

and−4+ x2 +2y2

xy2 (−4+ x2 + y2)√(4− x2 − y2)

= 0.

Therefore, x2 +2y2 = 4 and 2x2 + y2 = 4. Thus x = y and so x = y = 2√3. It follows from

the equation for z that z = 2√3

also. How do you know this is not the largest tetrahedron?

Example 23.1.7 An open box is to contain 32 cubic feet. Find the dimensions which willresult in the least surface area.

Let the height of the box be z and the length and width be x and y respectively. Thenxyz = 32 and so z = 32/xy. The total area is xy+ 2xz+ 2yz and so in terms of the twovariables x and y, the area is A = xy+ 64

y + 64x . To find best dimensions you note these must

result in a local minimum.

Ax =yx2 −64

x2 = 0, Ay =xy2 −64

y2 .

Therefore, yx2 − 64 = 0 and xy2 − 64 = 0 so xy2 = yx2. For sure the answer excludes thecase where any of the variables equals zero. Therefore, x = y and so x = 4 = y. Then z = 2from the requirement that xyz = 32. How do you know this gives the least surface area?Why is this not the largest surface area?

488 CHAPTER 23. OPTIMIZATIONExample 23.1.6 Find the volume of the smallest tetrahedron made up of the coordinateplanes in the first octant and a plane which is tangent to the sphere x7 +y? +22 =4.The normal to the sphere at a point (xo, yo, Zo) of the sphere is(x00 \/ 4-35 -)and so the equation of the tangent plane at this point issi 0) +3019 y0) + 4a —98 (2 433-98) ~042_ 2(4-9-9)Therefore, the function to minimize is. When z=0=y, x= 4, and when z=x=0, y= 4.When x = y=0, z= yofey) =3-—OS 6 xy/(4=8 = 7)This is because in beginning calculus it was shown that the volume of a pyramid is 1/3 thearea of the base times the height. Therefore, you simply need to find the gradient of thisand set it equal to zero. Thus upon taking the partial derivatives, you need to have—442x°+y?wy (4427 +y?) (4-2? —y?)=0,and44x74 2y*xy? (—4+2x° +y?) /(4—x? —y?)Therefore, x? + 2y* = 4 and 2x* + y? = 4. Thus x = y and sox = y= wat It follows from2V3the equation for z that z = ~~ also. How do you know this is not the largest tetrahedron?Example 23.1.7 An open box is to contain 32 cubic feet. Find the dimensions which willresult in the least surface area.Let the height of the box be z and the length and width be x and y respectively. Thenxyz = 32 and so z = 32/xy. The total area is xy + 2xz+ 2yz and so in terms of the twovariables x and y, the area is A= xy+ “ + o To find best dimensions you note these mustresult in a local minimum.Ay= =0, Ay =Therefore, yx* — 64 = 0 and xy* — 64 = 0 so xy” = yx*. For sure the answer excludes thecase where any of the variables equals zero. Therefore, x = y and sox =4= y. Then z= 2from the requirement that xyz = 32. How do you know this gives the least surface area?Why is this not the largest surface area?