Question

Justin wants to use 188 ft of fencing to fence off the greatest possible rectangular area for a garden. What dimensions should he use? What will be the area of the garden?

how do I solve this?

Answer 1

the greatest rectangular area would actually be a square

 so 188/4 = 47 feet per side

so 47ft x 47ft

area = 47^2 = 2209 square feet

Answer 2

Answer:

length =47 ft

width = 47 ft

Area=2209 square feet

Step-by-step explanation:

Hello

let's remember the  rectangular area equation

A=length (l)* width( w)

the perimeter  of that area is given by

P=2l+2w

step 1

Let

x=length of the greatest posible area

y=width of the greatest posible area

so

$A_{max} =x*y(equation\ (1)$

the perimeter of that area is

$P=2x+2y$

step 2

justin use 188ft,hence

$P=2x+2y=188\\2x+2y=188\\subtract\ 2x\ in\ each\ side\\2x+2y-2x=188-2x\\2y=188-2x\\\\divide\ each\ by\ 2\\\frac{2y}{2}=\frac{188-2x}{2} \\y=94-x$

y=94-x   equation (2)

step 2

Now replace (2) in (1)

$A_{max} =x*y\\A_{max} =x*(94-x)\\A_{max} =94x-x^{2}$

Now, he have the  area as a function of x

$A_{max} =94x-x^{2}$

derive to find the maxims of the function,by doing A(x)' = 0

$A_{max} =94x-x^{2}\\A' =94-2x \\A' =94-2x\\94-2x=0\\x=47$

x=47 (equation 3)

to figure out if is a maxim verify

$A(x)'' > 0?$

$A' =94-2x\\A'' =94-2\\92 >0$

so, x=94  

effectively is a maxim

Step 3

replace (3) in (2)

$A_{max} =x*y\\A_{max} =94x-x^{2}equation(4)\\replacing x=47 in (4)\\A_{max}=94*47-47^{2}\\ A_{max}=2209\ ft^{2}$

Step 4

now, replace (3) in (1) to find y

$y=94-x\\y=94-47\\y=47$

step 5

answer

the dimensions should be

length =47 ft

width = 47 ft

the greatest area possible is 47ft*47 ft = 2209 square feet

Have a great day