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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 $lw=24$ $w=\frac{24}{l}$




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

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

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Does the banner read Don't Panic? – Will Jagy Mar 9 '13 at 22:59
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
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 $l > \sqrt{18}$.

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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

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