Help with a question (1 Viewer)

Robin Hood · Nov 18, 2008

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ontVertAlignInTxbx/> <w:Word11KerningPairs/> <w:CachedColBalance/> </w:Compatibility> <w:BrowserLevel>MicrosoftInternetExplorer4</w:BrowserLevel> <m:mathPr> <m:mathFont m:val="Cambria Math"/> <m:brkBin m:val="before"/> <m:brkBinSub m:val="--"/> <m:smallFrac m:val="off"/> <m:dispDef/> <m:lMargin m:val="0"/> <m:rMargin m:val="0"/> <m:defJc m:val="centerGroup"/> <m:wrapIndent m:val="1440"/> <m:intLim m:val="subSup"/> <m:naryLim m:val="undOvr"/> </m:mathPr></w:WordDocument> </xml><![endif]--><style>  </style> A piece of wire 10m long is broken into two parts, which are bent into the shape of a rectangle and a square as shown. Find the dimensions x and y that make the total area a maximum.
(i dont have the pictures but if you just imagine a square shown with al sides labelled 'x' and a rectangle with the short sides labelled 'x' and the longer sides labelled 'y')
Can somebody help me with this? Can you show full working out please?

3unitz · Nov 18, 2008

total length 10m equals the perimeters of the square and rectangle:

10 = 6x + 2y ------(1)

total area is given by:

A_total = x^2 + xy

sub (1):
A_total(x) = x^2 + x(5 - 3x)
= 5x - 2x^2

we now differentiate to find the value of x for which A_total(x) is a maximum:

A_total'(x) = 5 - 4x

for maximum equate A_total'(x) to 0:
0 = 5 - 4x
ie. when x = 5/4 m

we know this is a maximum (and not minimum) as A_total''(x) = -4 which is negative

sub (1) to find y:
y = 5/4 m

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Robin Hood

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3unitz

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