Quadratic Equations

Question

A rectangular garden has dimensions of 23 feet by 10 feet. A gravel path of equal width is to be built around the garden. How wide can the path be if there is enough gravel for 234 square feet?

Answer 1

Let the width be $w$.

On the $23'$ sides, we'll have two rectangles of dimensions $23\cdot w$.

Similarly, on the $10'$ sides we'll have two rectangles of width $10\cdot w$.

In each corner, we'll have a $w\cdot w$, or $w^2$ dimension square.

Thus, our equation is this:

$$4\cdot w^2+2\cdot23w+2\cdot10w=234$$

Solving:

$$4w^2+66w-234=0$$

$$w=\frac{-66\pm\sqrt{4356-(4)(4)(-234)}}{2(4)}$$

Take positive answer since dimensions can't be negative.

$$w=\frac{-66+90}{8}$$

$$w=3'$$

Answer 2

You have two concentric rectangles.   The exterior of the given dimensions $23\times 10$, and an interior one of dimetions $(23-2x)\times(10-2x)$ for some path width $x$.   The difference in area is to be $234 \text{square.feet}$ of gravel path $x$ wide edging the garden.

Solve for $x$ using some quadratic equation described by the above.

Note: You will have two algebraic solutions, though one of which will be physically preposterous and the other a design folly (the path almost fills the area available in the garden).

Answer 3

Use the quadratic formula to solve the quadratic equation ax^2 +66x -234 =0, where a is the constant pi. The reason you do not have two concentric rectangles is that the outside "rectangle" has four quarter circles of radius x for its corners. I think a better way to phrase the problem is to say that a gravel path of constant width is to be build around the garden. If you have opened a garden gate of width x, you will have seen it sweeps out a sector of a circle of radius x.