Critical area

Calculus Level 4

Let ( A x , A y ) (A_x, A_y) be a point on the curve y = e x y=e^x such that the area of the region bounded by the y y -axis, the line y = A y , y=A_y, and the curve y = e x y = e^{x} is 1 0 6 . 10^6.

Find A x A_x to 2 decimal places.


The answer is 11.47.

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1 solution

I got a partial solution and then used numerical analysis to approximate a root (so if anyone can find another way I'd love to hear it).

Seeing as we want to minimise the total area and any area under the curve doesn't count towards critical area A y = e A x A_{y}=e^{A_{x}} (i.e. the point lies on the curve)

This leads to the conclusion that A x A y = A x e A x = t o t a l a r e a = 1 0 6 + 0 A x ( e x ) d x A_{x}A_{y}=A_{x}e^{A_{x}}=total area=10^{6}+\int_{0}^{A_{x}}(e^{x})dx = critical area + area under curve.

This rearranges to ( A x 1 ) e A x = 1 0 6 (A_{x}-1)e^{A_{x}}=10^6 and taking logs: A x = 6 l o g ( 10 ) l o g ( A x 1 ) A_{x}=6log(10)-log(A_{x}-1) .

I used this as an iteration formula with A x 1 = 10 A_{x_{1}}=10 (lucky guess) but if anyone has an easier way to solve this it would be great to hear it.

Thank You William for sharing the solution. The first time I wrote this problem I stated as a hint to use any method at one's disposal. Thus, if that method is a lucky guess, so be it. However, my solution relied on programming since I am quite a CS enthusiast. Therefore, I made short program of mine to approximate the solution, where the error of calculation would be less than 1 0 6 10^{-6}

Milan Milanic - 4 years, 3 months ago

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I used python for the last stage of the calculation (just ran 100 iterations of the formula which gave a very small error).

It's a really cool problem, I was just wondering if there was a way to solve the equation exactly.

William Whitehouse - 4 years, 3 months ago

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The exact answer can use the Lanbert-W function.

( x 1 ) e x = k (x-1)e^x=k ( x 1 ) e x 1 = k e (x-1)e^{x-1}=\frac{k}{e} x 1 = W ( k e ) x-1=W(\frac{k}{e}) x = 1 + W ( 1 0 6 e ) x=1+W(\frac{10^6}{e})

Finding its value on the other hand is another problem. I used an online calculator for that.

Shaun Leong - 4 years, 3 months ago

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