1. ## equation solving

I was able to answer parts (a) and (b) , however struggled with part (c). I've given the questions to part (a) and (b), as i think there is material given that is needed, in order to answer part (c)....

A rectangle has legth (l) and width (w).

(a) Write an expression for the perimeter and area in terms of l and w.

(b) If the perimeter is 100 meters and the area is 75 meters squared, change the expression from part (a) into equations

(c) Solve the equations to find the dimensions of the rectangle correct to 3 decimal places.

Thanks

2. a)

perimeter of rectangle = 2(length) + 2(width) = 2(l+w)

area of rectangular = length x width = lw

b)
given perimeter = 100 → 100 = 2(l+w) ------(1)
given area = 75 → 75 = lw -------- (2)

c)
from equation (2)
lw = 75
l = 75/w ------(3)

substitute (3) into (1)
100 = 2(75/w +w)
50w = (75 +w^2)
w^2 - 50w + 75 = 0
w = 48.4520788 or 1.547921201

substituting w into (3)
l = 1.547920201 or 48.4520788

3. Originally Posted by steph_r
I was able to answer parts (a) and (b) , however struggled with part (c). I've given the questions to part (a) and (b), as i think there is material given that is needed, in order to answer part (c)....

A rectangle has legth (l) and width (w).

(a) Write an expression for the perimeter and area in terms of l and w.

(b) If the perimeter is 100 meters and the area is 75 meters squared, change the expression from part (a) into equations

(c) Solve the equations to find the dimensions of the rectangle correct to 3 decimal places.

Thanks

(c)

From the Area ($\displaystyle 75 = lw$), we can rearrange to get:

$\displaystyle l=\frac{75}{w}$

Then you can substitute this into the perimeter equation ($\displaystyle 100 = 2(l+w)$):

$\displaystyle 2\left(\frac{75}{w}+w\right) = 100$

$\displaystyle \frac{75}{w}+w=50$

$\displaystyle w^2 - 50w + 75=0$

Since the discriminant is not a square number, i.e. $\displaystyle b^2 - 4ac = 2200$, you can't solve by trial and error.

Completing the square:

$\displaystyle w^2 - 50w + 625 - 625 + 75 =0$

$\displaystyle (w-25)^2 =550$

$\displaystyle w-25 = \pm \sqrt{550}$

$\displaystyle w-25 =\pm 5\sqrt{22}$

$\displaystyle w = 25 \pm 5 \sqrt{22}$

Then the Length is (via rationalizing the denominator)

$\displaystyle l = \frac{75}{25 \pm 5 \sqrt{22}}=\frac{75}{25 \pm 5\sqrt{22}} \times \frac{25 \mp 5 \sqrt{22}}{25 \mp 5 \sqrt{22}}=\frac{75(25 \mp 5\sqrt{22})}{625 - 25 \times 22}=\frac{75(25 \mp 5 \sqrt{22})}{75}=$

$\displaystyle l=25 \mp 5\sqrt{22}$

So when the width is $\displaystyle 25 + 5\sqrt{22}$, the length is $\displaystyle 25-5\sqrt{22}$ and when the length is $\displaystyle 25 - 5 \sqrt{22}$, the width is $\displaystyle 25 + 5 \sqrt{22}$.

However... the length is always longer than the width, so your final solution is

Length = $\displaystyle 25 +5\sqrt{22}=48.452\ \mbox{m}$

Width = $\displaystyle 25 - 5\sqrt{22}=1.548\ \mbox{m}$