Q36. - Geometry

Question

A farmer has100m of fencing with which with to make a rectangular corral. A side of a barn will be used as one side of the corral, as shown in the overhead view.

a. If the width of the corral is x , express the length and the area in terms of x.

Step-by-Step Solution

Answer

$\begin{array}{l}Area:A=(100-2x)x\\ Length:l=100-2x\end{array}$

Step 1. Given information.

A farmer has100mof fencing with which to make a rectangular corral. A side of a barn will be used as one side of the corral, as shown in the overhead view.

. Step 2. Concept Used.

The perimeter of the rectangle is $P=2(length+breadth)$

The area of the square is$A={\left(side\right)}^2$

Step 3. Let’s find the area and length.

Perimeter of the fence 

$\begin{array}{l}l+x+x=100\\ 2x+l=100\\ l=100-2x\\ now\\ A=lb\\ A=(100-2x)x\end{array}$

Thus 

Area:$A=(100-2x)x$

Length:$l=100-2x$

1. Make a graph showing values of x on the horizontal axis and the corresponding areas on the vertical axis.

Step 1. Given information.

A farmer has100mof fencing with which to make a rectangular corral. A side of a barn will be used as one side of the corral, as shown in the overhead view.

. Step 2. Concept Used.

The area of the rectangle is 
$A=length\times breadth$

Step 3. Let’s graph.

For different values of x finding area.

The table shows the Area for different value of x.

$A=(100-2x)x$

The Graph is shown below.

1. What dimensions give the corral the greatest possible area?

Step 1. Given information.

A farmer has100m of fencing with which to make a rectangular corral. A side of a barn will be used as one side of the corral, as shown in the overhead view.

Step 2. Concept Used.

The area of the rectangle is 
$A=length\times breadth$

Step 3. Let’s find the dimensions of the corral that gives the maximum area.

From the table it is seen that 

When $\begin{array}{l}x=25\\ A=(100-2x)x\\ A=(100-50)50\\ A=2500\end{array}$

Which is greatest area.

Therefore the dimensions of the corral that gives the maximum area: $50\times 25$