Question

A farmer wishes to fence in a rectangular field of 10,000 feet square. The north south fences cost RM1.50 per foot
while the cast west fences will cost RM6.00 per foot. Find the dimensions of the field that will minimize the cost.[x=50,y=200]​

Answer 1

Answer:

the cost of fencing is rupees

3567

hope it helps

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Answer 2

Question :-

A farmer wishes to fence in a rectangular field of 10,000 feet square. The north south fences cost Rs 1.50 per feet

while the east west fences will cost Rs 6.00 per foot. Find the dimensions of the field that will minimize the cost.

Answer

Given :-

• Area of Rectangular feild = 10000 sq. feet
• Cost of North and South Fencing is Rs 1.50 per feet.
• Cost of west Fencing is Rs 6.00 per feet.

To find :-

• Dimensions of rectangle that minimize the cost.

$\begin{gathered}\Large{\bold{\pink{\underline{CaLcUlAtIoN\::}}}} \end{gathered}$

$\begin{gathered}\bf\red{Let} \end{gathered}$

$\begin{gathered}\longmapsto\:\:\bf{Breadth\:(b) \:is \:y\:feet}. \\ \end{gathered}$

$\begin{gathered}\longmapsto\:\:\bf{Length\:(l) \:is \:x\:feet}. \end{gathered}$

$\longmapsto\tt\boxed{\bf \:Area\:of\:Rectangle=l\times{b}}$

$\bf \:⟼ \: 10000 = x \times y$

$\bf \:⟼ \: y = \dfrac{10000}{x} \: ⟼ \: (1)$

$\longmapsto\tt\boxed{\bf \:Perimeter\:of \: Rectangle=2(l + b)}$

$\bf \:⟼P = 2(x + y)$

Cost of fencing the rectangular feild, C

$\bf \: ⟼ C = 1.5 \times x + 1.5 \times x + 6 \times y + 6 \times y$

$\bf \: ⟼ C = 3x + 12y$

On substituting the value of y from equation (1), we get

$\bf \: ⟼ C = 3x + 12 \times \dfrac{10000}{x}$

$\bf \: ⟼ C = 3x + \dfrac{120000}{x}$

⟼ Differentiate w. r. t. x, we get

$\bf \: ⟼ \dfrac{d}{dx} C = \dfrac{d}{dx} (3x + \dfrac{120000}{x} )$

$\bf \: ⟼ \dfrac{dC}{dx} = 3 - \dfrac{120000}{ {x}^{2} } \: ⟼ \: (2)$

For maximum or minimum value,

$\bf \: ⟼ \dfrac{dC}{dx} = 0$

$\bf \: ⟼ 3 - \dfrac{120000}{ {x}^{2} } = 0$

$\bf \: ⟼ 3 {x}^{2} = 120000$

$\bf \: ⟼ {x}^{2} = 40000$

$\bf \: ⟼ x = 20 \: feet$

Differentiate (2) w. r. t. x, we get

$\bf \: ⟼ \dfrac{ {d}^{2} C}{d {x}^{2} } = 0 + \dfrac{240000}{ {x}^{3} } > 0$

$\bf \: ⟼ C \: is \: minimum \: at \: x = 200$

Put x = 200 in equation (1), we get

$\bf \: ⟼ y = \dfrac{10000}{200} = 50 \: feet$

$\begin{gathered}\Large\bf\red{Hence,} \\ \end{gathered}$

$\bf \: ⟼ ➣ Breadth \: is  \: 50 \: feet$

$\begin{gathered}\bf\red{And,} \\ \end{gathered}$

$\bf \: ⟼ ➣ Length \: is \: 200 \: feet.$