Question

Question #63e9a

Answer 1

`4y+2x=80`
`4y=80-2x`
`y=20-1/2x`

The length is determine by the y coordinate of the point in the first quadrant and the width is determined by the x coordinate.

The area of the rectangle is length by width.

`therefore` the area `=x*(20-1/2x)`

Let A = Area

`A=20x-1/2x^2`

By deriving the Area function, a rate of change of area can be found.

`A^'=20-(1/2*2)x`
`A^'=20-x`

The stationary point, and hence when the area is either at a maximum or minimum value is found when the rate of change is zero.

`therefore 0=20-x`
`x=20`

To ensure that this value is indeed a maximum, find the second derivative. If the second derivative is a negative value, at that point, then the value is a maximum.

`A^"''"=-1`

Since the second derivate is negative the area is at a maximum at `x=20`

Substitute this x value into the original equation

`y=20-1/2x`
`y=20-1/2*20`
`y=20-10`
`y=10`

The dimension of the rectangle to produce the maximum area are:

x=20 and y=10