Max/min problem (1 Viewer)

[kpad5991]

The diagram in the link below represents in metres the dimensions of a small garden.
Show that y= (20-x^2).

https://imgur.com/gallery/BDt5l

 

[Queenroot]

Bump

 

[Sp3ctre]

I'm on holidays atm so I don't have any writing equipment to work out this question, but try splitting the trapezium into a right angle triangle and a square and equate the areas.

i.e. Area of trapezium = Area of square + Area of rectangle

Then rearrange so that y is the subject

 

[fan96]

I got y² = 20 - x²

(Did I do something wrong?)

I'm on holidays atm so I don't have any writing equipment to work out this question, but try splitting the trapezium into a right angle triangle and a square and equate the areas.

i.e. Area of trapezium = Area of square + Area of rectangle

Then rearrange so that y is the subject

Unfortunately that doesn't work. From that, you get:

(1/2x)(x) [rectangle] + 1/2xy [triangle] = 1/2x(x+y) [trapezium]
i.e. 1/2(x+y) = 1/2(x+y)

 

[Sp3ctre]

I got y² = 20 - x²

(Did I do something wrong?)

View attachment 34362



Unfortunately that doesn't work. From that, you get:

(1/2x)(x) [rectangle] + 1/2xy [triangle] = 1/2x(x+y) [trapezium]
i.e. 1/2(x+y) = 1/2(x+y)

Oh of course, I'm stupid D:

The working you provided seems correct, perhaps OP may have made a typo