Question

A rectangular garden of area 456 square feet is to be surrounded on three sides by a brick wall costing $ 12 per foot and on one side by a fence costing $ 7 per foot. Find the dimensions of the garden such that the cost of the materials is minimized.

Answer 1

Answer:

$l=2\sqrt{114}feet$    $b=2\sqrt{114}feet$      $\left ( l,b \right )=(21.3541,21.3541)$

Step-by-step explanation:

Given,

Area of garden=456 square feet

Brick wall  cost three sides $=\ 12$ per foot

fencing cost of one side$=\ 7$ per foot

Let length and breadth of rectangular field is          (l,b) feet

For minimizing the cost of material perimeter of garden should be minimum and area of garden is more.

Perimeter of garden$=2\left ( l+b \right )=k$

$l+b=k/2$

$l=k/2-b$

A=$l\times b=456$

A=$\left ( k/2-b \right )b=456$

Differentiating the equation to find maximum area

$\frac{d}{db}(\left ( k/2-b \right )b)=\frac{\partial 456}{\partial b}$

$k/2-2b=0$

$k=4b$

$l=4b/2-b$

$l=b$

To find $maxima$  again differentiating

$\frac{\partial A^2 }{\partial b^2}=\frac{\partial \left ( k/2-2b \right )}{\partial b}<0$

$\frac{\partial A^2 }{\partial b^2}<0=-2$

The area is maximum when    l=b

and perimeter is minimum

$l\times b=456$

$b^{2}=456$

$b=2\sqrt{114}feet$

$l=2\sqrt{114}feet$

 Cost material

$=2\sqrt{114}\times 3\times 12+2\sqrt{114}\times 7$

$=72\sqrt{114}+14\sqrt{114}$

$=\ 86\sqrt{114}$

$=\ 918.22873$

Dimension of rectangular garden

$\left ( l,b \right )=(21.3541,21.3541)$