Question

Melissa is building a rectangular area for her cow, Daisy. She plans to build the area along a straight stretch of river (no fence is needed along the river).
a. Melissa would like to give daisy 800m^2 of grazing room. What is the least amount of fencing she will need?

b. She realizes that she only has 60 m of fencing. What is the largest grazing area she will be able to give daisy?

Answer 1

least amount of fencing needed = 20 +20+40 = 80 m

largest grazing area by 60m fencing = 250m²

pls mark as brainliest

Answer 2

$\large\underline\purple{\bold{Solution - a }}$

Let us assume that

• Length of Rectangular area = 'x' m

and

• Breadth of Rectangular area = 'y' m

So,

• According to statement,

$\longmapsto\bf \: Area_{Rectangular)} \: = 800 \: {m}^{2}$

$\rm :\implies\:xy = 800$

$\rm :\implies\:x = \dfrac{800}{y} - - (1)$

Now,

• Length of fencing, L = x + 2y

So,

$\rm :\implies\:L \: = \: \dfrac{800}{y} + 2y$

Differentiating both sides w. r. t. y, we get

$\rm :\implies\:\dfrac{dL}{dy} = - \dfrac{800}{ {y}^{2} } + 2 - - - (2)$

For maxima or minima,

$\longmapsto \red{ \bf \:\dfrac{dL}{dy} = 0 }$

$\rm :\implies\: - \dfrac{800}{ {y}^{2} } + 2 = 0$

$\rm :\implies\:2 = \dfrac{800}{ {y}^{2} }$

$\rm :\implies\: {y}^{2} = 400$

$\rm :\implies\: \boxed{ \blue{ \bf \: y \: = \: 20 \: m}} - - (3)$

Now,

Again differentiating (2) w. r. t. y, we get

$\rm :\implies\:\dfrac{ {d}^{2} L}{ {dy}^{2} } = \dfrac{1600}{ {y}^{3} } > 0$

$\rm :\implies\:\:\boxed{ \pink{\bf\:L \: is \: minimum \: when \: y \: = \: 20 \: m}}$

Now,

Substituting the value of y in equation (1), we get

$\rm :\implies\:x \: = \: \dfrac{800}{20}$

$\rm :\implies\: \boxed{ \green{ \bf \: x\: = \: 40 \: m}}$

Hence,

The fence is shortest if the side parallel to the river has length 40 m and the other 2 sides each have length 20 m.

$\: \boxed{ \purple{ \rm \: Least \: Length \: of \: fencing \: = 40 + 20 + 20 = 80 \: m}}$

$\large\underline\purple{\bold{Solution - b}}$

Let us assume that

• Length of Rectangular area = 'x' m

and

• Breadth of Rectangular area = 'y' m

So,

• According to statement,

Length of fencing = 60 m

$\rm :\implies\:x + 2y = 60$

$\rm :\implies\:x = 60 - 2y - - (1)$

Now,

$\longmapsto \red{ \bf \: Area_{Rectangular)} = xy}$

$\rm :\implies\:A = (60 - 2y)y$

$\rm :\implies\:A = 60y - 2 {y}^{2}$

Differentiating both sides w. r. t. y, we get

$\rm :\implies\:\dfrac{dA}{dy} = 60 - 4y - - (2)$

For, maxima or minima,

$\bf \longmapsto \: \bf \:\dfrac{dA}{dy} \: =\: 0$

$\rm :\implies\:60 - 4y = 0$

$\rm :\implies\:60 = 4y$

$\rm :\implies\:\:\boxed{\pink{\bf\:y \: = \: 15 \: m}} - - (3)$

Now, again differentiating both sides w. r. t. y, we get

$\rm :\implies\:\dfrac{ {d}^{2}A }{ {dy}^{2} } = - 4 < 0$

Hence, Area is maximum.

$\rm :\implies\:\:\boxed{ \pink{\bf\:A \: is \: maximum \: when \: y \: = \: 15 \: m}}$

So,

Substituting the value of y from equation (3) in (1), we get

$\rm :\implies\:x = 60 - 2 \times 15$

$\rm :\implies\: \boxed{ \green{ \bf \: x \: = \: 30 \: m}}$

Hence,

$\:\:\boxed{ \pink{\bf\:Largest \:Area \: is \:xy = 30 \times 15 = 450 \: {m}^{2} }}$