Question

A​ landscaper, who just completed a rectangular flower garden measuring 15 feet in length by 7 feet in​ width, orders 1 cubic yard of premixed​ cement, all of which is to be used to create a border of uniform width around the garden. if the border is to have a depth of 44 ​inches, how wide will the border​ be?

Answer 1

The width of the border will be 0.1648... ft  or  1.9786... inches.

Explanation

Lets assume, the width of the border is  $x$ ft.

The rectangular garden is 15 ft in length and 7 ft in width. So, the area of the garden $=(15*7)ft^2= 105 ft^2$

Now, the length of the garden including border $=(15+2x) ft$ and the width of the garden including border $=(7+2x) ft$

So, the area of the garden including the border $= (15+2x)(7+2x) ft^2$

Thus the area of the border $=[(15+2x)(7+2x)-105] ft^2$

Total amount of premixed cement needed is  1 yard³ = 27 ft³

As the depth of the border is 44 inches or $(\frac{44}{12})ft$ or $(\frac{11}{3})ft$, so....

$\frac{11}{3}[(15+2x)(7+2x)-105] = 27\\ \\ 11(105+44x+4x^2 -105)=81\\ \\ 11(4x^2+44x)=81\\ \\ 44x^2+484x-81=0$

Using quadratic formula....

$x= \frac{-484+/-\sqrt{484^2-4(44)(-81)}}{2(44)}\\ \\ x= \frac{-484+/-\sqrt{234256+14256}}{88}\\ \\ x= \frac{-484+/-\sqrt{248512} }{88}\\ \\ x= 0.1648...$

(Negative value ignored)

So, the width of the border $=0.1648...ft= (0.1648...*12)inches= 1.9786...inches$