Optimization: triangular prism

yhpigs

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Aug 12, 2018
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I have a question that I'm really stuck on at the moment.

A chocolate factory will sell a new chocolate bar with a weight of 100g.

They will sell for $2.00 each and 300 000 will be sold each week. For every price rise of 10c, sales will drop by 5 100 bars per week.

Manufacturing cost is 50c/100g of chocolate, + 0.213 cents/cm^2 for the wrapping.

The shape of the chocolate bar is an equilateral triangular prism.



We have to:

a) find an expression for 'h' (perpendicular height of the prism) in terms of 'b' (base length)

b) fins expression for 'A' (area of equilateral triangle) in terms of 'b' [would this just be A=1/2 bh?]

c) find equation for surface area in terms of b

d) find the dimensions of the chocolate bar.



I'm really confused on this and have no idea where to start. Any help would be greatly appreciated
 
… find an expression for 'h' (perpendicular height of [an equilateral triangle]) in terms of 'b' (base length) …
Do you remember trigonometry?

Half of an equilateral triangle (with base b) is a right triangle.

The right triangle's base is 1/2∙b

It's height is h

Draw and label a picture.

You know the three interior angles of an equilateral triangle are all the same (that's the definition of an equilateral triangle). You know these three angle measures sum to 180°, so you know each interior angle is 60°.

Therefore, in your right-triangle picture, h (height) is the side opposite a 60° angle and 1/2∙b (base) is the side adjacent to that angle.

From the right-triangle definitions of trigonometric functions, we define the tangent function as:

tan(angle) = Opposite/Adjacent

tan(60°) = h/[1/2∙b]

Evaluate tan(60°), and then solve for h. Questions? :cool:
 
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… [find] expression for 'A' (area of equilateral triangle) in terms of 'b' [would this just be A=1/2 bh?] …
Yes! But substitute your expression for h, and simplify, to get the area in terms of b. :)
 
Do you remember trigonometry?

Half of an equilateral triangle (with base b) is a right triangle.

The right triangle's base is 1/2∙b

It's height is h

Draw and label a picture.

You know the three interior angles of an equilateral triangle are all the same (that's the definition of an equilateral triangle). You know these three angle measures sum to 180°, so you know each interior angle is 60°.

Therefore, in your right-triangle picture, h (height) is the side opposite a 60° angle and 1/2∙b (base) is the side adjacent to that angle.

From the right-triangle definitions of trigonometric functions, we define the tangent function as:

tan(angle) = Opposite/Adjacent

tan(60°) = h/[1/2∙b]

Evaluate tan(60°), and then solve for h. Questions? :cool:

Thank you for the help!! So to solve for h, would the equation be: h=tan 60(1/2∙b) ? This is what I got solving for h.
 
… h=tan 60(1/2∙b) ? This is what I got solving for h.
Yes, but your notation needs fixing because (outside of context) it could be read as the tangent of angle [60(1/2∙b)]°

When typing functions, I like to use function notation:

h = tan(60°)∙1/2∙b

or

h = b/2 tan(60°)

PS: Write a degree symbol, if you're not expressing radians. (Angle measures without a degree symbol denote radians.)
 
If my equation for h is right, this should be A=1/2b(0.32 * 1/2∙b) (tan 60 0.32)
0.32 is not accurate for tan(60°)

Your calculator is set to radian mode, instead of degree mode.

Does your class use decimal approximations, in general? Using rounded approximations, instead of exact values, tends to introduce error. After several steps of rounding, the total error may lead to an incorrect end result. Using exact values also helps to simplify expressions.

You're expected to memorize sine and cosine for some special angles. (Knowing sine and cosine is sufficient to calculate tangent because tangent=sine/cosine.)

TrigSpecialAngles.JPG
 
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Does your class use decimal approximations, in general? Using rounded approximations, instead of exact values, introduces error. After several steps of rounding, the total error leads to an incorrect end result. Using exact values also helps to simplify expressions.

Ah right, we did learn exact values! Thank you~
 
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