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.

100m by 112.5m

Solutions would be much appreciated!
Attached Images
 Untitled picture.jpg (19.9 KB, 9 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: 18,953 Thanks: 1599 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: 407

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: $\:9x + 8y \:=\:1800\;\;[1]$

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

Solve equation [1] for $y$.
$\quad y \;=\;\dfrac{1800-9x}{8} \quad\Rightarrow\quad y \;=\;225 - \tfrac{9}{8}x\;\;[2]$

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

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 $\quad y \;=\;\dfrac{1800-9x}{8} \quad\Rightarrow\quad y \;=\;225 - \tfrac{9}{8}x\;\;[2]$ 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 [2]: $\:y \:=\:225 - \frac{9}{8}(100) \:=\: \dfrac{225}{2}$ 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.

 Thread Tools Display Modes Linear 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