# Practical use of Calculus

• Jul 23rd 2006, 08:42 PM
NeF
Practical use of Calculus
15.1

A rectangular container has the following dimensions (cm)

Length $\displaystyle 3x$
Width $\displaystyle 90-3x$
depth $\displaystyle \frac{x}{3}$

Determine
15.1.1 Volume V in terms of $\displaystyle x$
15.1.2 The value for $\displaystyle x$ 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 $\displaystyle 2x^3-96x+2304$

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!
• Jul 23rd 2006, 10:22 PM
earboth
Quote:

Originally Posted by NeF
15.1
A rectangular container has the following dimensions (cm)
Length $\displaystyle 3x$
Width $\displaystyle 90-3x$
depth $\displaystyle \frac{x}{3}$

Determine
15.1.1 Volume V in terms of $\displaystyle x$
15.1.2 The value for $\displaystyle x$ 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 $\displaystyle 2x^3-96x+2304$
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,

$\displaystyle V=l\cdot w\cdot d \Longrightarrow V(x)=3x\cdot (90-x)\cdot \frac{x}{3}=-3x^3+90x^2$

You'll get the maximum value for V if $\displaystyle \frac{dV}{dx}=0$. Thus:

$\displaystyle \frac{dV}{dx}=-9x^2+180x$. Therefore: $\displaystyle -9x^2+180x=0$.
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: $\displaystyle x^2+(48-x)^2=x^2+2304-96x+x^2=2x^2-96x+2304$. I assume that there isa typo in your text.

As you can see $\displaystyle s(x)=2x^2-96x+2304$ is the equation of a parabola opening upward. So the smallest value is at the vertex where the derivative of s is zero:

$\displaystyle \frac{ds}{dx}=4x-96$. this expression is zero if x = 24. That means if x = y = 24 you get the smallest sum of the squares.

Greetings

EB
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}