Question

Doris has a fish pond 15 by 18 ft. She wants to put grass in a strip of uniform with around the pond. She has enough grass seed for 234 ft squared. How wide will the strip be? Round your answer to the nearest tenth, when appropriate

Answer 1

3 feet

Explanation:

Let x be the width of the grass strip.

So the length of the grass strip=18+2x

Area of the grass strip =2(18+2x)(x)+2(15)(x)=2(18x+2x^2)+2(15x)=(36x+4x^2)+30x=4x^2+66x^2

But the available grass seed is enough for 234 ft squared

4x^2+66x=234

4x^2+66x-234=0

x={-66+-[(66)^2–4(4)(-234)]^1/2}/2(4)

x={-66+-[4356+3744]^1/2}/8={-66+-[8100]^1/2}/8

x={-66+-(90)}/8

Two values of x: x=(-66+90)/8=24/8=3 and the other value of x=(-66–90)/8=-19.5. Adopt the positive value of x=3 feet (answer)

So the width of the strip is 3 feet.

Check for the area of grass strip around the fish pond : Area=2(18+6)(3)+2(15)(3)=2(24)(3)+90=144+90=234 square feet

Other area check =4x^2+66x=4(3)^2+66(3)=36+198=234 square feet

Therefore the strip is 3 feet wide

Answer 2

Check the picture below.

so if we simply subtract the strip area with the pond and the area of the pond only, what's leftover is the area of the strip alone, so

`\stackrel{\textit{area of the grass \underline{with} pond}}{(15+2x)(18+2x)}-\stackrel{\textit{area of the pond}}{(15)(18)}=\stackrel{\textit{their difference}}{232} \\\\\\ 270+30x+16x+4x^2-270=232\implies 46x+4x^2=232 \\\\\\ 23x+2x^2=116\implies 2x^2+23x-116=0`

`~~~~~~~~~~~~\textit{quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{2}x^2\stackrel{\stackrel{b}{\downarrow }}{+23}x\stackrel{\stackrel{c}{\downarrow }}{-116}=0 \qquad \qquad x= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ x = \cfrac{-23\pm √(23^2-4(2)(-116))}{2(2)}\implies x = \cfrac{-23\pm √(1457)}{4}`

now, the negative root will simply yield a negative value, which is not applicable here, so we'll only use the positive root, since it's the only feasible one for the phenomena

`x = \cfrac{-23+ √(1457)}{4}\implies x \approx 3.792667341996157\implies \stackrel{\textit{rounded up}}{x = 3.8}`