Question

a rectangular dog kennel is to be constructed alongside a house with 60m of fencing. If the house serves as one side of the kennel, determine the greatest possible area that can be enclosed.

Answer 1

Answer:

$450\text{ m}^2$.

Step-by-step explanation:

Let x represent side of kennel opposite to house and y represent other sides.

We have been given that a rectangular dog kennel is to be constructed alongside a house with 60 m of fencing.

Since fencing will cover 3 sides of kennel, so perimeter of kennel would be:

$x+y+y=60$

$x+2y=60$

Let us solve for x.

$x+2y-2y=60-2y$

$x=60-2y$

The area of the kennel would be product of its sides that is:

$\text{Area}=x\cdot y$

Now, we will substitute $x=60-2y$ in area equation as:

$A(y)=(60-2y)\cdot y$

$A(y)=60y-2y^2$

Let us find the derivative as shown below:

$A'(y)=60-4y$

Now, we will set derivative equal to 0 and solve for y.

$60-4y=0$

$60=4y$

$\frac{60}{4}=\frac{4y}{4}$

$y=15$

Upon substituting $y=15$ in area function, we will greatest possible area.

$A(y)=60y-2y^2$

$A(15)=60(15)-2(15)^2$

$A(15)=900-2(225)$

$A(15)=900-450$

$A(y)=450$

Therefore, the greatest possible area that can be enclosed by 60 m of fencing is 450 square meters.