Question

Prove that of all rectangular parllclopiped of the same volume. The cube has the least surface. Find the minimum value of x²+y²+z² given that ax+by+cz. Find the minimum value of x²+y²+z² when yz+zx+xy=3a².

Answer 1

The minimum value of x² + y² + z² for the expression ax + by + cz, subject to the constraint yz + zx + xy = 3a², is 6a².

Step-by-step explanation:

Rectangular parallelepiped is a three-dimensional figure with six rectangular faces. To prove that among all rectangular parallelepipeds with the same volume, a cube has the least surface area, consider a rectangular parallelepiped with side lengths a, b, and c. The surface area (S) is given by the formula:

S = 2(ab + bc + ca)

Now, for a cube with side length s, the volume (V) is s³ and the surface area is6s² . For the given parallelepiped, let abc = V. Using the inequality
`\(2(ab + bc + ca) \geq 6\sqrt[3]{a^2b^2c^2} = 6\sqrt[3]{V^2}\),` we can show that the cube has the least surface area.

Moving on to the second question, to find the minimum value of x² + y² + z² for the expression ax + by + cz under the constraint yz + zx + xy = 3a², we can use Lagrange multipliers. The Lagrangian function is:

`\[ L(x, y, z, \lambda) = x² + y² + z² + \lambda(3a² - yz - zx - xy) \]`

Taking partial derivatives and setting them to zero, we get a system of equations. Solving this system leads to the values x = ±a, y = ±a, z = ±a, and λ = ±1. Substituting these values back into the expression x² + y² + z², we find that the minimum value is 6a².

In conclusion, the minimum value of x² + y² + z² under the given constraint is 6a², achieved when x = ±a, y = ±a, and z = ±a.