# The management of the UNICO department store has decided to enclose an 862 ft^2 area outside the...

## Question:

The management of the UNICO department store has decided to enclose an 862 ft{eq}^2 {/eq} area outside the building for displaying potted plants and flowers. One side will be formed by the external wall of the store, two sides will be constructed of pine boards, and the fourth side will be made of galvanized steel fencing. If the pine board fencing costs $4/running foot and the steel fencing costs $2/running foot, determine the dimensions of the enclosure that can be erected at minimum cost.

## Area Of Rectangle

The rectangular area is defined as the product of its sides that are:

(1) Length.

(2) Breadth.

The front view of a matchbox has a rectangular shape. The standard unit of area is meter square.

## Answer and Explanation:

**Given**

- Wood side is {eq}x\;{\rm{ft}} {/eq}

- Steel side is {eq}y\;{\rm{ft}} {/eq}

The expression for the area is given by

{eq}\begin{align*} {\rm{A}} &= x \times y\\ 862 &= x \times y\\ x &= \dfrac{{862}}{y}......\left( 1 \right) \end{align*} {/eq}

The expression for the cost,

{eq}\begin{align*} {\rm{C}} &= 2x \times 4 + y \times 2\\ {\rm{C}} &= 8x + 2y \end{align*} {/eq}

Put in value of {eq}y {/eq} in above equation

{eq}{\rm{C}} = 8x + 2 \times \dfrac{{862}}{x} {/eq}

Differentiate above equation,

{eq}{\rm{C'}}\left( {\rm{x}} \right) = 8 - \dfrac{{862 \times 2}}{{{x^2}}}......\left( 2 \right) {/eq}

Put {eq}{\rm{C'}}\left( {\rm{x}} \right) = 0 {/eq}

{eq}\begin{align*} 8 &= \dfrac{{862 \times 2}}{{{x^2}}}\\ {x^2} &= \dfrac{{1724}}{8} \end{align*} {/eq}

Differentiate equation (2)

{eq}{\rm{C''}}\left( {\rm{x}} \right) = 0 + 862 \times 2 \times \dfrac{2}{{{x^3}}} {/eq}

When {eq}x = \sqrt {\dfrac{{1724}}{8}} {/eq} , then {eq}{\rm{C''}}\left( {\rm{x}} \right) > 0 {/eq}

Therefore, cost is minimum when {eq}x = \sqrt {\dfrac{{1724}}{8}} = 14.67 {/eq}

Put the value of {eq}x {/eq} in equation (1)

{eq}y = 58.75 {/eq}

Therefore, wood side is {eq}14.67\;{\rm{ft}} {/eq}

And, steel side is {eq}58.75\;{\rm{ft}} {/eq}

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from Geometry: High School

Chapter 8 / Lesson 7