User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

August 17th, 2014, 05:39 AM   #1
Newbie

Joined: Jul 2014
From: australia

Posts: 16
Thanks: 0

Question is attached:

don't understand part c.

Answer for part c is:

100m by 112.5m

Solutions would be much appreciated! Attached Images Untitled picture.jpg (19.9 KB, 10 views)

Last edited by skipjack; August 17th, 2014 at 06:23 AM. August 17th, 2014, 06:28 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,373 Thanks: 2010 You need to assume that all of the available fencing is used. For part (c), they want you to calculate x and y. August 17th, 2014, 08:57 PM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Hello, jessjans11!

Quote:
 5. 1800 m of fencing is available to enclose 6 identical pig pens. Code:  * - - * - - * - - * | | | | y * - - * - - * - - * | | | | y * - - * - - * - - * x x x (a) Explain why $9x+8y \:=\:1800.$

There are 9 lengths of fencing which are $x$ m each.
There are 8 lengths of fencing which are $y$ m each.

The total fencing is 1800 m: $\:9x + 8y \:=\:1800\;\;$

Quote:
 (b) Show that the area of each pen $\quad$is given by: $\,A \:=\:-\tfrac{9}{8}x^2 + 225x$

Solve equation  for $y$.
$\quad y \;=\;\dfrac{1800-9x}{8} \quad\Rightarrow\quad y \;=\;225 - \tfrac{9}{8}x\;\;$

The area of a pen is: $\:A \;=\;xy \;=\;x\left(225-\frac{9}{8}x\right)$

Therefore: $\:A \;=\;-\frac{9}{8}x^2 + 225$

Quote:
 (c) If the enclosed area is to be a maximum, $\quad$what are the dimensions of each pen?

We want to maximize $A$.

The equation is a down-opening parabola.
Its maximum is at its vertex.

The vertex is: $\:x \:=\:\dfrac{\text{-}b}{2a} \:=\:\dfrac{\text{-}225}{2(\text{-}\frac{9}{8})} \:=\:100$

Substitute into : $\:y \:=\:225 - \frac{9}{8}(100) \:=\: \dfrac{225}{2}$

Therefore, the dimensions are: $\:100\text{m} \times 112.5\text{m}$. August 18th, 2014, 06:19 AM   #4
Banned Camp

Joined: Jun 2014
From: Earth

Posts: 945
Thanks: 191

Quote:
 Originally Posted by soroban $\quad y \;=\;\dfrac{1800-9x}{8} \quad\Rightarrow\quad y \;=\;225 - \tfrac{9}{8}x\;\;$ The area of a pen is: $\:A \;=\;xy \;=\;x\left(225-\frac{9}{8}x\right)$ Therefore: $\:A \;=\;-\frac{9}{8}x^2 + 225$ Substitute into : $\:y \:=\:225 - \frac{9}{8}(100) \:=\: \dfrac{225}{2}$ Therefore, the dimensions are: $\:100\text{m} \times 112.5\text{m}$.
Your A expression is incorrect. You left the x off of "225x."

And then when you substituted, you substituted into the wrong expression.

You should have substituted the x-value into

A = 225x - (9/8)x^2. Tags quadratics Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post grangeeducation Algebra 1 April 10th, 2014 09:12 AM NeverForgetVivistee Algebra 3 November 21st, 2011 04:35 PM manich44 Algebra 7 November 5th, 2011 02:21 AM sallyyy Algebra 3 June 4th, 2011 01:58 AM maria186 Algebra 1 February 10th, 2010 05:34 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      