Optimizing Area and Volume: Solving Maximum Problems

[Dr Zoidburg]

[SOLVED] Maximum area/volume problems

I have two here. One I feel I've solved but am just looking for reassurance (I'm needy that way!), The other I've got an answer but I know I've missed something.

Homework Statement

1. A rectangle is inscribed in a 6/8/10 cm right-angle triangle where two vertices (and one side) touch the hypotenuse side and the other two vertices each touching one of the other two sides.
Find the greatest area posssible.

2.A closed rectangular container with a square base is to have a volume of 2000cm$$^{3}$$. It costs twice as much per cm$$^{2}$$ for the top and bottom as it does for the sides. Find the dimensions of least cost.

The Attempt at a Solution

For the first one, I made the hypotenuse the base of the triangle. I then worked out the height of the triangle using trig (height = 4.8cm)
I then used the equivalent triangles theory to state that the base: height ratio of the 6/8/10 triangle is 25:12 (10:4.8), therefore the same ratio would apply for the triangle made within by the rectangle, at the top of the large triangle (hope that's clear!).
i.e. If we take the height of the smaller inner triangle to be h and the base b, then the ratio of b:h is 25:12, making b=25h/12. This is one side of the rectangle.
The height of the rectangle is 4.8-h, making the area b*(4.8-h) = 25h/12(4.8-h)
f(h) = 10h - 25h$$^{2}$$/12
f'(h) = 10 - 25h/6
putting f'(h) to 0, we get h=2.4, making b=5 and the area 24cm$$^{2}$$.
Now assuming this is correct, I was curiously wondering if there's any other way of doing this. I justa sort of madea this uppa as I went.

2. base of box = x, height = h.
Volume of box = x$$^{2}$$h = 2000, h = 2000/$$x^{2}$$
Surface area of box = 2x$$^{2}$$ + 4xh
SA = 2x$$^{2}$$ + 4x.2000/$$x^{2}$$ = 2x$$^{2}$$ + 8000/x
f'(SA) = 4x - 8000/$$x^{2}$$ = 0
4$$x^{3}$$ - 8000 = 0
$$x^{3}$$ = 2000
x = 12.6, which means h also = 12.6
This is correct in every aspect except it ignores the extra bit of the cost of the box sides/base. Where/how do I bring that info into the equation?

 

[HallsofIvy]

I have two here. One I feel I've solved but am just looking for reassurance (I'm needy that way!), The other I've got an answer but I know I've missed something.

Homework Statement

1. A rectangle is inscribed in a 6/8/10 cm right-angle triangle where two vertices (and one side) touch the hypotenuse side and the other two vertices each touching one of the other two sides.
Find the greatest area posssible.

2.A closed rectangular container with a square base is to have a volume of 2000cm$$^{3}$$. It costs twice as much per cm$$^{2}$$ for the top and bottom as it does for the sides. Find the dimensions of least cost.

The Attempt at a Solution

For the first one, I made the hypotenuse the base of the triangle. I then worked out the height of the triangle using trig (height = 4.8cm)
I then used the equivalent triangles theory to state that the base: height ratio of the 6/8/10 triangle is 25:12 (10:4.8), therefore the same ratio would apply for the triangle made within by the rectangle, at the top of the large triangle (hope that's clear!).
i.e. If we take the height of the smaller inner triangle to be h and the base b, then the ratio of b:h is 25:12, making b=25h/12. This is one side of the rectangle.
The height of the rectangle is 4.8-h, making the area b*(4.8-h) = 25h/12(4.8-h)
f(h) = 10h - 25h$$^{2}$$/12
f'(h) = 10 - 25h/6
putting f'(h) to 0, we get h=2.4, making b=5 and the area 24cm$$^{2}$$.
Now assuming this is correct, I was curiously wondering if there's any other way of doing this. I justa sort of madea this uppa as I went.

I started to do this by setting the right triangle on a coordinate system so that the 8 cm side was on the x- axis: so hypotenuse is a line with slope 6/8= 3/4, so the sides of the rectangle are lines with slope -4/3, ... until I read through your solution! It is much simpler to do the problem with similar triangles, but I think my method is just slightly different from yours. Again, imagine the triangle with the 8 cm side horizontal and 6 cm side vertical. Let h be the length of the side of the rectangle perpendicular to the hypotenus and let b be the length of the side of the rectangle in the hypotenuse. Also let x be the length of the hypotenuse above the rectangle (above b) and let y be the length of the hypotenuse below the rectangle (below b). You are correct that the triangles formed at each end are similar to the original triangle and so we have h/x= 8/6 or x= (3/4)h. Also h/y= 6/8 and so y= (4/3)h. Since x+ y+ b= 10, (3/4)h+ (4/3)h+ b= 10 giving b= 10- (25/12)h. That area is hb= 10h- (25/12)h² The derivative of that is 10- (25/6)h so h= 60/25= 12/5= 2.4 cm. Then b= 10- (25/12)(12/5)= 5 cm. However I feel I should point out that 5(2.4)= 12 square cm, not 24! Indeed, the entire triangle has area 24 square cm. so you certainly could not fit such a rectangle in the triangle.

2. base of box = x, height = h.
Volume of box = x$$^{2}$$h = 2000, h = 2000/$$x^{2}$$
Surface area of box = 2x$$^{2}$$ + 4xh
SA = 2x$$^{2}$$ + 4x.2000/$$x^{2}$$ = 2x$$^{2}$$ + 8000/x
f'(SA) = 4x - 8000/$$x^{2}$$ = 0
4$$x^{3}$$ - 8000 = 0
$$x^{3}$$ = 2000
x = 12.6, which means h also = 12.6
This is correct in every aspect except it ignores the extra bit of the cost of the box sides/base. Where/how do I bring that info into the equation?

The sides of the box have area xz and there are 4 of them: total area 4xz. The top and bottom of the box have area x²: total area 2x². Suppose the sides cost "a" per square cm. Then they cost a total of 4axz. The top and bottom cost twice that: 2a per square cm and so a total of 4ax². The total cost is 4axz+ 4ax² and that is what you must minimize. You should find that the "a" does not appear in the final answer. (And x is not 12.6. As you would guess, since the bottom and top cost more than the sides, the "cheapest" box has more sides, less to and bottom.)

 

[Dr Zoidburg]

D'oh!
I meant 12cm$$^{2}$$ not 24. thanks for spotting that error.
And thanks for the advice on Q2.

 

[Dr Zoidburg]

Okay. Here's what I got for the box question:
base of box = x, height = h.
Volume of box = x²h = 2000; therefore h = 2000/x²
Surface area of box = 2x² + 4xh

Assume the sides cost $a p/cm².
Then the total cost of the sides will be 4axh. Since the top and bottom cost twice as much (ie. 2a) than the sides, total cost for top & bottom = 2ax² + 2ax² = 4ax².

Total cost (TC) of box = 4axh + 4ax²
Since h = 2000/x², we can write this as:
TC = 4ax² + 8000a/x
We differentiate this for x, to find the maximum length of the base:
f’(x) = 8ax – 8000a/x²
put this to 0 and multiply both sides by x², we get:
8ax³ - 8000a = 0
8ax³ = 8000a
x³ = 1000
x = 10 cm
Volume of box = x²h = 2000
h = 2000/100
h = 20 cm

Dimensions of least cost of a 2000cm³ box where base is a square and top & bottom cost twice as much as the sides is to have the base a 10x10 cm square and the sides 20cm.

whew! did it, tg.