Optimum perimeter and area

Maria is planning to build a glass greenhouse. To have room for all the plants she wants, the greenhouse needs to be 36 square metres in size. To make it easy to build, she has chosen a rectangular shape. Glass is expensive, so she wants to use as little as possible. She wants the perimeter of glass around the greenhouse to be as small as possible.

This would be the best, the optimal thing to do. How do we do that? Let’s try a few numbers.. 12 metres times 3 metres gives an area of 36 square metres. The perimeter is then 12 plus 3 plus 12 plus 3, equals 30 metres.

But with 9 metres in length, a width of 4 metres is needed to get an area of 36 square meters. So the perimeter is then 9 plus 4 plus 9 plus 4, equals 26 metres. That’s a smaller perimeter, a more optimal result. How can we know when we get the best result, the smallest perimeter? To find out, we use letters, variables, for the length and the width.

If the length of the greenhouse is x and the width is y, then the formula for the area is: x times y. Maria wants a greenhouse that is exactly 36 square metres. She wants to constrain the area to this. x times y equals 36 is then the constraint equation. The perimeter is x plus y plus x plus y, equals 2x plus 2y.

Maria’s desired, optimal, result is to minimize the perimeter. The formula for the perimeter is the optimization equation. We use two unknown values, x and y, to calculate the perimeter. But it would be easier if we only had one unknown value. The formula can be changed so that it only contains one unknown value.

Let’s choose x, the length. Now back to the constraint equation: x times y equals 36 Divide both sides of the equation by x. y is then 36 divided by x. Then we replace y in the optimization equation with 36 divided by x. The perimeter is 2x plus 2y equals 2x plus 2 times 36 divided by x and 2 times 36 is 72.

Pause the video and check that’s correct. Now let’s make a table where we enter different values of the length of the greenhouse, x, and see what we get for the perimeter. In the first column, A, we enter the x-value. In the next column, B, we enter the formula for the perimeter: 2x plus 72 divided by x. The cheapest glass panes Maria has found have a width of 1 metre.

So let’s increase x by 1 each time in column A. If x equals 1, the perimeter is 2 times 1, plus 72 divided by 1, equals 74. Then we increase x to 2 and calculate the perimeter again. We see that the perimeter decreases as x increases. But when x becomes 7, the perimeter begins to increase.

Of all the values of x we examined, we get the smallest perimeter when x equals 6. Then the width, y, equals 36 divided by 6, which is 6. If Maria builds a greenhouse with a length of 6 metres and a width of 6 metres, she gets 36 square metres. The greenhouse should have the shape of a square. This will give the smallest perimeter, the optimal result for Maria.