Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Hello everyone how would the following word problem.

A rectangular banner has a red border and rectangular white center. The width of the border at top and bottom is 8 Inches and along the sides it is 6 inches, The total area of the whole banner is 24 square feet. What would be the dimensions height and width of the banner that maximize the area of the white center.

I know that 12 Inches equals 1 foot. l would be length and w width

so area would be $lw=24$ $w=\frac{24}{l}$

$A=(L-1)(w-\frac{4}{3})$

$A=\frac{76}{3}-\frac{24}{l}-\frac{4l}{3}$

$A'(l)=\frac{24}{l^2}-\frac{4}{3}$

for l I got $l=\sqrt{18} feet $

and for w width I got $w=5.67$ feet

share|improve this question
1  
Does the banner read Don't Panic? en.wikipedia.org/wiki/Towel_Day –  Will Jagy Mar 9 '13 at 22:59
2  
The calculations are essentially correct. You may not be using the calculator correcly, the width is closer to $5.657$. Perhaps the issue is rounding and rekeying. Need to show we have obtained the maximum. –  André Nicolas Mar 9 '13 at 22:59
    
Yes that makes sense –  Fernando Martinez Mar 9 '13 at 23:03

1 Answer 1

up vote 2 down vote accepted

Try to express $w = \dfrac{24}{l}$ in terms of $l = \sqrt{18} = 3\sqrt 2$, to get the precise value of $w$.

So $w = \dfrac{24}{l} = \dfrac{24}{3\sqrt 2} = \dfrac 8{\sqrt 2}$

And note that indeed, $A = w\cdot l = \dfrac{24}{3\sqrt 2}\cdot 3{\sqrt 2} = 24$

Using a calculator and rounding gets you a bit in the way of a loss of precision.

Lastly, you're calculation of the only possible solution to $A'(l) = 0$ is $l = \sqrt {18} = 3\sqrt 2$, since the alternative is negative, and since $l$ represents length, $l > 0$.

But even though the only place an extrema can occur in this case is at your solution $l$, and we can guess that since the problem is asking for maximization, that your solution must be a maximum, you really do need to show in your work that $A(\sqrt {18})$ is indeed the maximum value: Show that $A' > 0 $ on $0 < l < \sqrt{18}$, and $A' < 0$ for $l > \sqrt{18}$.

share|improve this answer
    
i see thanks for the help –  Fernando Martinez Mar 14 '13 at 18:08
    
Hi, @Babak ! Yes, I got to bed very late, and ended up sleeping until 10:00 a.m.! (3+1/2 hours ago). I've been "on" and "off" since then! How are you? –  amWhy Mar 14 '13 at 18:30
    
@amWhy: Good. :-) Thanks God, just exhausted of doing jobs around home today. –  Babak S. Mar 14 '13 at 18:33

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.