A cardboard box without a lid is to have a volume of 16,384 cm3. Find the dimensions that minimize the amount of cardboard used. (Let x, y, and z be the dimensions of the cardboard box.)

The dimensions that minimize the amount of cardboard used is

x = 31 cm ,  y = 34 cm & Z = 15.54 cm

Step-by-step explanation:

Volume of the cardboard = 16,384

The function that represents the area of the cardboard without a lid is given by

------  (1)

Volume of the cardboard with sides x, y & z is

Put this value of z in equation (1) we get

Differentiate above equation with respect to x & y we get

Take

 ------ (2)

 ------- (3)

By solving equation (2) & (3) we get

x = 31 cm

From equation 2

y = 32768 ()

y = 34 cm

Z = 15.54 cm

Thus the dimensions that minimize the amount of cardboard used is

x = 31 cm ,  y = 34 cm & Z = 15.54 cm

i am currently in the exact same situation rn. algebra2/adv algebra is killing

step-by-step explanation:

the degrees would remain the same ( the number to the left of the decimal point.

the multiply the decimal by 60 to get the minutes and use the numbers to the left of the decimal point.

then multiply the decimal portion of that by 60 for the seconds and round off as needed.

39 .956652

the 39 remains the degrees.

the minutes = 0.956652 x 60 = 57.39912

the minutes = 57

the seconds = 0.39912 x 60 = 23.95 = 24 ( rounded to nearest tenth)

so 39.956652 = 39 degrees, 57 minutes and 24 seconds

-75.159809

the -75 remains the degrees

the minutes = 0.159809 x 60 = 9.58854

the minutes = 9

the seconds = 0.58854 x 60 = 35.31, rounded to nearest tenth = 35.3 seconds

so -75.159809 = -75 degrees, 9 minutes and 35.3 seconds.

the first coordinate is north or south, since it it positive it is north.

the second number is east or west, because it is negative it is west.

39 degrees, 57 minutes and 24 seconds n

-75 degrees, 9 minutes and 35.3 seconds w