Question

Shenelle has 100100100 meters of fencing to build a rectangular garden. the garden's area (in square meters) as a function of the garden's width www (in meters) is modeled by: a(w)=-(w-25)^2+625a(w)=−(w−25) 2 +625a, left parenthesis, w, right parenthesis, equals, minus, left parenthesis, w, minus, 25, right parenthesis, start superscript, 2, end superscript, plus, 625 what is the maximum area possible?

Answer 1

Answer:

625 sq. m.

Step-by-step explanation:

The graden's area is given as :  $a(w) = -(w-25)^2+625$

For maximization of area,we need to put first order derivative of given equation equals to 0 i.e. a'(w) should be equal to zero.  

$a'(w) = -2(w-25) = 0$

On Simplifying :

$w-25 = 0$

w = 25 meters

Since we are given that Shenelle has 100 meters of fencing to build a rectangular garden

This means perimeter is equal to 100.

Formula of perimeter of rectangle : $2(w+l) = 100$

Where l is length

w is width

Plugging w as 25

$2(25+l) = 100$

Dividing 2 in both sides

$(25+l) = 50$

$l = 25$

So, for the maximum area the length should be 25 m and width should 25 m

So, So, maximum area $=Length \times Width = 25 \times 25 = 625 m^2$

Hence the maximum possible area is 625 sq. m.

Answer 2

A function assigns the values. The maximum area possible of Shenelle's field is 625 meters².

What is a Function?

A function assigns the value of each element of one set to the other specific element of another set.

To find the width of the garden that will give the maximum area, the function of the area is needed to be differentiated with respect to width(w). Therefore,

$a(w)=-(w-25)^2+625\\\\a(w)= -w^2-625+50w+625\\\\a(w)=-w^2+50w\\\\\dfrac{da}{dw} = \dfrac{d(-w^2+50w)}{dw}\\\\a'= -2w+50$

Equate it with 0 to know the value of w,

$-2w+50=0\\\\2w=50\\\\w=25$

Thus, the area of the rectangular garden will be maximum when the width of the rectangular garden will be 25 meters.

Substitute w=25 in the function of the area to get the maximum area of the rectangular field,

$a(w)=-(w-25)^2+625\\\\a(25)=-(25-25)^2+625\\\\a(25)=625\ meters^2$

Hence,  the maximum area possible of Shenelle's field is 625 meters².

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