A rectangular box without a lid is to be made from 48 m2 of cardboard. Find the maximum volume of such a box. SOLUTION We let x, y, and z to be the length, width, and height, respectively, of the box in meters. Then we wish to maximize V = xyz subject to the constraint g(x, y, z) = 2xz + 2yz + xy = 48.

(1) = λ                     (2) = λ                  (3) = λ

yz = λ (2z + y)              xz = λ (2z + x)              xy = λ (2x + 2y)

From (1) and (2) => (λx)(2z) + λxy = (λy)(2z) + xyλ

                         => x = y

x² = λ(4x)

x = 4λ which is also = y   ;     4λz = λ(2z + 4λ)

                                                     => 2z = 4λ  ∴ z = 2λ    

zxy + 2yz + xy = 48

=> λ = 1        => x = 4 = y      &     z = 2

= (4)(4)(2) = 32

      at (x,y,z) = (4,4,2)

Step-by-step explanation:

-7/8

step-by-step explanation:

instead of subtracting you add the -6/8 and -1/8

well what you have to do is divide 84 by 6 so theanswer

step-by-step explanation:

84 divided by 6 = 14

and that lightbulb changing party must be freaking lit!