# Johnny went for a drive with Sara

By now, Johnny’s mom trusted his driving and readily agreed to his borrowing the car for a short drive in the countryside with his girlfriend Sara.

The weather was nice – no clouds in sight and the temperature also felt just right – not too cold or not a high temperature that would fry an egg on the road. So they opened the car windows to enjoy the breeze while driving. Johnny had not decided where to go and Sara didn’t mind that. Being together for a drive in the countryside was all she wanted.

# Farmer selling apples on the roadside

While cruising they saw a farmer selling apples on the roadside. He had bushels of apples and a chair to sit on. The apples were visibly inviting. They bought a few. The first bite told Johnny that the apples were delicious. So Sara also started nibbling on them.

Johnny started chatting with the farmer. While talking, the farmer realized that these were two educated kids and told them of his dilemma.

Farmer: I don’t have a large farm and it is difficult to make a living.

Johnny: How many apples does your farm produce each year?

Farmer: I have 50 trees and each tree produces about 800 apples.

Sara: That would be 40,000 apples a year. I guess it is hard to live on the small income from them.

Johnny: Why can’t you plant more trees? Do the trees have to be as far apart as they are now?

# How many apple trees can the farmer add to the orchard?

Farmer: I asked my father the same question. He talked to other orchard owners. He said that for every tree you add, you will decrease the yield of all the trees by 10 apples per tree. I don’t know what to do. I am not well educated. I didn’t even go to high school.

Sara: Johnny, that’s easy. Use your calculus.

Johnny: How is this a calculus problem?

Sara: Remember, when we were doing slopes of y = sin x, we found the slope to be zero when the curve was at a plateau. The total apple production curve based on the number of trees will also plateau at the point of optimum production and the derivative will be zero.

Johnny: But the derivative will be zero at maximum or minimum of the curve.

Sara: Let’s figure out where the curve plateaus. Then we can check if it is for optimum production or for a disaster.

Johnny: He has 50 trees. If he adds x trees, the total production will be the number of trees (50 + x) multiplied by the yield per tree (800 – 10x).

Sara: So the production will be 40000 +300 x -10 x^{2}.

Johnny: The first derivative will be 0 + 300 – 20x.

Sara: So the answer is that x will be optimum when 300 – 20x = 0 or x =15.

Johnny: That’s all good and dandy but how do you know that this is not the answer to the minimum production but to the maximum one?

Sara: Good, we should always check that. If he plants 15 more trees he will have a total of 65 of them. The production from each tree will be 800 minus 10 times 15 = 650. The farm would then produce 42,250 apples. That’s more than the current production. Therefore, it is not a minima but the optimum value.

So, they told the farmer that as per their calculation, the best deal is to plant 15 more trees in the farm. That would give him 2500 more apples each year.

Farmer: Okay, I will try that. If you are right, I owe you a free bushel of apples.

Johnny was happy with that offer, and off they went.

# Graphical solution

At home, Sara made this graph on her lap top. It was based on the formula that they had figured out. Indeed, the graph peaked at 15 extra trees planted. Johnny liked that the answer was verified by the graph. Of course, he also realized that using calculus they were able to get the answer very fast.

Suddenly, Johnny asked Sara how at the farm she could multiply 65 by 650 in a jiffy without using a calculator. Sara just smiled.

Later Sara told him that another way to check whether the plateau is maxima or minima is from the sign of the second derivative. A positive second derivative indicates a minima and a negative one means it is a maxima. For example, here the production P = 40000 +300 x -10 x^{2}. The plateau occurred at dP/dx = 300 -20x. The second derivative d^{2}P/dx^{2} = -20 indicating the first derivative solution gave a maxima. Also they had verified the plateau not to be a minima by other calculations.

*Challenge*

Josie has construction paper which is 48 cm^{2 }in area. She wants to make a square base box open at the top with the walls perpendicular to it to get a maximum volume. She is wondering what the box would look like and what its volume would be. Sara knows what to tell her. Do you?

* Solution: *Let us start by saying that the box has a square base with each side being x cm, and has a height of y cm.

The surface area A = x^{2} + 4xy

Given A = 48, and then x^{2} + 4xy = 48 or x + 4y = 48/x or y = 12/x – x/4

The volume of the box V = x^{2}y or V = x^{2 }(12/x -x/4) or V= 12x -x^{3}/4.

At maxima/minima, dV/dx has to be zero.

For

V= 12x -x^{3}/4, dV/dx = 0 = 12 – 3x^{2}/4 or 3x^{2} = 48 or x^{2} = 16 or x = ± 4.

Negative length of the base does not make sense and hence x = 4.

We know that the area A = x^{2} + 4xy = 48. With x =4, 16 +16y=48 or 16y =32 or y =2 cm

So the volume of the box V = x^{2}y = 4^{2} x 2 = 32 cm^{3}.

We know that a box with a zero height will have a zero volume. The value of 32 cm^{3 }is greater and hence the plateau is a maxima and not a minima.