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The width of a rectangle is five less than twice the length. What is the minimum area of the rectangle?

Sagot :

W=2L-5

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A=L*W

=L(2L-5)

=2L²-5L

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dA/dL=4L-5

When dA/dL=0,

4L-5=0

4L=5

L=5/4

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When L=5/4

A=5/4*(2*(5/4)-5)

=5/4*(10/4-20/4)

=5/4*((-10)/4)

=5/4*(-5/2)

=-25/8

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