Area Optimization Problem
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Area Optimization Problem

[From: ] [author: ] [Date: 12-06-25] [Hit: ]
Labeling the triangles side x and the squares side y,P = 3x + 4y,An equation for the total area would be the sum of the individual areas.I divided the equilateral triangle into two halves. Each angle is 60 deg.I solved the perimeter equation for y to get y = (10 - 3x)/4.......
I have no idea what I'm doing wrong here, I was so sure I was doing everything correctly but the numbers in the end aren't making sense. Could anyone tell me what I'm doing wrong?

"The sum of the perimeters of an equilateral triangle and a square is 10. Find the dimensions of both shapes that produce a minimum total area."

Labeling the triangle's side 'x' and the square's side 'y', the perimeter's equation is

P = 3x + 4y, which has been constrained to be 10 = 3x + 4y

An equation for the total area would be the sum of the individual areas.

A = y^2 (square's) + 1/2*x*h (triangle)

I divided the equilateral triangle into two halves. Each angle is 60 deg. Simple trig:

sin(60) = h/x
h = xsin(60)

So my area equation is:

A = y^2 + 1/2*x^2*sin(60)

I solved the perimeter equation for y to get y = (10 - 3x)/4.

Plugging this in to y^2 and simplifying a bit:

A = (100 - 60x - 9x^2)/16 + sq.rt(3)/4*x^2

> A = 100/16 - 60x/16 - 9x^2/16 + sq.rt(3)/4*x^2

Took derivative:

dA/dx = (0 - 60/16 - 18x/16) + sq.rt(3)/2*x

Set dA/dx to zero to start finding relative maxima

0 = -60/16 - 18x/16 + x*sq.rt(3)/2

...which gives me roots that are negative, which is impossible for the problem I'm currently doing.

I've looked this over and over and can't find what I'm doing wrong. Is there something conceptually I'm missing?

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The following line is wrong from your answer:

A = (100 - 60x - 9x^2)/16 + sq.rt(3)/4*x^2

It should be:

A = (100 - 60x + 9x^2)/16 + sqrt(3)/4*x^2

You simply have a sign error. Hope that helps.
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