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# In a right prism with triangular base, given the sum of the areas of three mutually adjacent faces...?

Sorry, full question: In a right prism with triangular base, given the sum of the areas of three mutually adjacent faces (that is, of two lateral faces and one base) is 24, find the maximum volume of the prism.

Please help!! Thanks so much! :D

### 2 Answers

- MorewoodLv 79 years agoFavorite Answer
Vekkus4 is on the right track. For an optimization problem:

0) Determine the variables.

1) Write a formula (in n variables) for the function to be optimized: F(x,y,z,w)

2) Write a formula for the constraint: G(x,y,z,w) = constant

3) Use Lagrange Multipliers to get an additional n equations in a total of (n+1) unknowns.

∇F(x,y,z,w) = λ•∇G(x,y,z,w)

4) Solve the system of equations from (2) & (3):

(n+1) equations in (n+1) unknowns, usually not linear!

5) Plug each solution into F(x,y,z,w) to get best value (in this case the maximum).

However, Vekkus4's hypothesis of an equilateral base produces a maximum volume of less than 10.4 (and requires solving a cubic). Using more variables (for a more general triangular prism) produces not only simpler equations, but a maximum more than 50% larger.

Based on earlier questions, DetectiveKat has been doing Lagrange Multipliers for weeks, so that part shouldn't be the problem. Deciding on variables and getting the formulas (which are intimately related problems) must be the difficulty. Since the area formula for the rectangular sides of the prism are very simple, and the volume is also just base times height, perhaps the difficulty lies with the area of the triangular base. Here are some area formulas to try: http://www.cliffsnotes.com/study_guide/Areas-of-Tr... There are several approaches, which is the easiest?

Alternatively, since I like linear algebra, consider a more general triangular prism:

Place the vertex shared by the three adjacent faces at the origin (0,0,0).

Turn the prism to place the base containing that vertex in the xy-plane.

Rotate the prism around the z-axis to put a second vertex on the x-axis: X=(a,0,0).

Then the third vertex of that base takes the form: Y=(b,c,0)

while the final vertex connected to the:origin is of the form: Z=(d,e,h).

{For a right prism we would require d=e=0.}

Now the volume is just the triple product: |½(X×Y)•Z|, half the matrix determinant:

| a 0 0 |

| b c 0 |

| d e h |

while the areas of 3 adjacent sides are cross-products: ½|X×Y|+|Y×Z|+|Z×X|=24

Notice that, even though we have six variables (seven with λ) the equations are reasonably nice since three of the partial derivatives of the Volume are zero. Since, for the maximum volume, ½(ach)≠0, one can quickly deduce the values for b, d, and e using just those three equations. {First solve ∂G/∂d=0 for "d", use that result in ∂G/∂b=0 to get "b" (and also "d"), and finally ∂G/∂e=0 gives "e".} Substituting these values leaves four equations which become linear in each variable (they are still quadratic equations but the square roots vanish) and which are straight-forward to solve.

For non-calculus students, one could make geometric arguements, involving shearing, for why the three shared edges of those three sides must be mutually perpendicular. For three mutually perpendicular faces, this volume is just the square root of half the product of the three faces:

V = √[ABC/2] 24 = A + B + C.

So we need to maximize the product of three numbers whose sum is 24. Replacing any two numbers with their average leaves the sum unchanged but the product (hence the volume) larger:

A=D+ε, B=D-ε ⇒ AB=D²-ε² ≤ D² while A+B = (D+ε) + (D-ε) = D+D

Hence the maximum volume cannot be achieved unless all three faces have the same area. So what three equal numbers add to 24? And what is the square root of half their product? Done!

This last example brings us back to the crux of the original problem - choosing variables. An unconventional but interesting choice of variables would be the three areas (A>0 and B>0 the rectangular areas, and C>0 the triangular area) along with the angle between the rectangular sides, call that θ. (Since the prism is "right" the angles between rectangular and triangular sides are both 90°.) Then the volume is:

F( A, B, C, θ) = √( ABCsinθ / 2 )

with G( A, B, C, θ ) = A + B + C = 48

and Lagrange gives us the additional equations:

λ = √[ BCsinθ / (8A) ]

λ = √[ ACsinθ / (8B) ]

λ = √[ ABsinθ / (8C) ]

0 = cosθ √[ ABC / (8sinθ) ]

Where the last immediately implies (since A,B,C>0) that θ=90° and the previous three imply that A=B=C, leading quickly to the maximum possible volume.

- vekkus4Lv 69 years ago
Cool I love calc problems that are stated in a minimum of words.

Step 1. Figure out the shape of a triangular right prism with maximum volume

According to some principles or experience, the triangle is equilateral.

(I might get that proved below)

Step 2. Therefore the only two variables are s (the side of the equilateral triangle) and the height h

Step 3. Write down the constraint

area of base = √3/2 s^2

area of lateral side = s h

total of two lateral sides and base: √3/2 s^2 + 2 s h = 24

Step 4. Write down the function to be maximized

V(s,h) = √3/2 s^2 h

Step 5. Do you know Lagrange Multipliers? Then use Lagrange Multipliers!

Otherwise substitute the constraint into the function to be maximized, then maximize it as a function of one variable. In Lagrange Multipliers basically the gradient of the function to be maximized (the "objective" function, in optimization algorithm talk) is orthogonal to the surface described by the constraint. To do the substitution, h = (24 - √3/2 s^2)/(2s)

so V(s) = √3/2 s^2 (24 - √3/2 s^2)/(2s)

and then you maximize that with respect to s and plug that back in to get h.