Thread: Practical use of Calculus

15.1

A rectangular container has the following dimensions (cm)

Length
Width
depth

Determine
15.1.1 Volume V in terms of
15.1.2 The value for for which it has a max volume

15.2 The sum of two positive numbers x and y is 48
15.2.1 find y in terms of x
15.2.2 Prove that the sum of thier squares can be given by

15.2.3 Find the samllest possible value of the sum of thier squares

I have no idea >< I made a couple of attempts but it doesnt look right. weh!

Originally Posted by NeF

15.1
A rectangular container has the following dimensions (cm)
Length
Width
depth

Determine
15.1.1 Volume V in terms of
15.1.2 The value for for which it has a max volume
15.2 The sum of two positive numbers x and y is 48
15.2.1 find y in terms of x
15.2.2 Prove that the sum of thier squares can be given by
15.2.3 Find the samllest possible value of the sum of thier squares
I have no idea >< I made a couple of attempts but it doesnt look right. weh!

Hello,NeF,



You'll get the maximum value for V if . Thus:

. Therefore: .
Solve this equation for x and you'll get x = 0 or x = 20. x = 0 produces obviously the minimum value of V, so at x = 20 V has its maximum.

to 15.2:
x + y = 48. Therefore:
y = 48-x

Sum of squares: . I assume that there isa typo in your text.

As you can see is the equation of a parabola opening upward. So the smallest value is at the vertex where the derivative of s is zero:

. this expression is zero if x = 24. That means if x = y = 24 you get the smallest sum of the squares.

Greetings

EB