# Are 105 ft of panel good to enclose a rectangular area of 1400 square feet if a wall is on one side?

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What is the area of 105 ft. panel? If you tell me the area in square feet or square meter or the dimension of length and width of the panel, Then I will answer your question ... ok...

• The idea seems to be that you will put up three panels, and use the existing wall for the 4th side of your enclosed area.

Let x be the length of each of the two parallel panels that you will erect.

Then 105 - 2x is the length of the third panel, and also the enclosed length of the existing wall that you are going to use.

Therefore the enclosed area is A = x * (105-2x) = 105x - 2x^2

dA/dX = 105-4x

Maximum area at 4x=105 or x=26 1/4 ft

So the maximum-area enclosure will be 26 1/4 x 52.5 ft. This is 1378 ft^2

You cannot enclose 1400 ft^2 (though you can come close)

• To my feeble brain, the 105 feet of panel will be used to make three sides of the enclosure, and one third of 105 feet of wall will be used as the fourth.

A square will give the greatest area, so it will have 33 feet on a side. This will result in an area of 1089 sq ft.

• you can use differential calculus to do it!

just get two set of eqn's

1. perimeter = 105 ft, so 105ft = 2x + 2y

where y and x represents length and width , respectively.

2.area =1400 sq ft, so x* y =1400.

use this two eqn's to get single eqn without y and than, you can find gradeint function and at a turning point, it's zero so you get values of turning points.use it to determine max and min , and you could use second derivative here.

than, you could find the max and min values of area and see whether it gonna cover 1400sq ft.

OR you could guestimate it!

Source(s): i do maths