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icycledough

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Let me know if I've got the right answer, but it should go something like this.

So we can call the 2 remaining sides x and 8 - x

Area = x (8 - x) = 8x - x^2

We want to differentiate this in terms of x to find the maximum

dA / dx = 8 - 2x

dA/dx = 0 when x = 4

To prove it is a maximum, we can find the 2nd derivative; that would just be -2 (as this is negative, there is a maximum when x = 4)

So the dimensions would be 4 metres x 4 metres (x and 8 - x, when x = 4)
 

lolzlolz

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Let me know if I've got the right answer, but it should go something like this.

So we can call the 2 remaining sides x and 8 - x

Area = x (8 - x) = 8x - x^2

We want to differentiate this in terms of x to find the maximum

dA / dx = 8 - 2x

dA/dx = 0 when x = 4

To prove it is a maximum, we can find the 2nd derivative; that would just be -2 (as this is negative, there is a maximum when x = 4)

So the dimensions would be 4 metres x 4 metres (x and 8 - x, when x = 4)
why is the reamining sides x and 8-x
 

icycledough

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It mentions that there is 8 metres of garden edging for the border on the other 2 sides, so I assumed that it meant the other 2 sides have a total length of 8 metres, thus I made them x and 8 - x
 

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