Cube optimization, trouble with formula

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mickey222

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Here is the problem: The design for a cube is being put together which will have a volume of 17,000 cubic inches. Both the top and the bottom are square, and the material for the top and bottom will cost 4 cents per square inch. Each side is made of a different material which will cost 7 cents per square inch. The square top and bottom of the cube is fused to the sides with adhesive material which costs 2 cents per linear inch. The sides are fused with each other with adhesive that costs 5 cents per linear inch.

1. What dimensions of the cube will incur the least cost?
2. Determine the least cost.

bhj.png

I begin with V (volume)=17,000
since volume for this cube will equal x^2h, I write: 17,000=x^2h

So I have my volume formula, now to determine the cost formula.

C=7(4xh) + 4(2x^2)

reasoning: the sides are rectangular, each square inch costing 7 cents, so 7 times 4xh (because there are 4 sides). The top/bottom are square, each square costing 4 cents, so 4(2x^2), because there is 1 top and 1 bottom.

Now to incorporate the cost of the adhesive onto this. I came up with (in red):

C=7(4xh)+4(2x^2)+2(8x)+(5(4h)).

Reasoning: if each side of the top and bottom are fused with adhesive costing 2 cents each, and there are four sides on the top, and four on the bottom, then 2(8x). Now, the rectangular sides are all fused with one another, meeting at each corner of the cube, and the adhesive costs 5 cents per linear inch, so 5(4h).

But its a complete mess... simplified, I get C= 28xh+8x^2+16x+20h.
But now I have to plug H in, which is equal to 17,000/x^2 (from our original volume equation).

The result is an even bigger mess...

Have I gone wrong somewhere?
 
mickey222 said:
Here is the problem: The design for a cube is being put together which will have a volume of 17,000 cubic inches. Both the top and the bottom are square, and the material for the top and bottom will cost 4 cents per square inch. Each side is made of a different material which will cost 7 cents per square inch. The square top and bottom of the cube is fused to the sides with adhesive material which costs 2 cents per linear inch. The sides are fused with each other with adhesive that costs 5 cents per linear inch.

1. What dimensions of the cube will incur the least cost?
2. Determine the least cost.

View attachment 25140

I begin with V (volume)=17,000
since volume for this cube will equal x^2h, I write: 17,000=x^2h

So I have my volume formula, now to determine the cost formula.

C=7(4xh) + 4(2x^2)

reasoning: the sides are rectangular, each square inch costing 7 cents, so 7 times 4xh (because there are 4 sides). The top/bottom are square, each square costing 4 cents, so 4(2x^2), because there is 1 top and 1 bottom.

Now to incorporate the cost of the adhesive onto this. I came up with (in red):

C=7(4xh)+4(2x^2)+2(8x)+(5(4h)).

Reasoning: if each side of the top and bottom are fused with adhesive costing 2 cents each, and there are four sides on the top, and four on the bottom, then 2(8x). Now, the rectangular sides are all fused with one another, meeting at each corner of the cube, and the adhesive costs 5 cents per linear inch, so 5(4h).

But its a complete mess... simplified, I get C= 28xh+8x^2+16x+20h.
But now I have to plug H in, which is equal to 17,000/x^2 (from our original volume equation).

The result is an even bigger mess...

Have I gone wrong somewhere?
Click to expand...
It all looks good to me, given that "cube" is a euphemism for "square-based prism" or equivalent.

Do the next steps you stated, and you'll end up having to solve a fourth-degree equation, which may require technology. But that last bit is the only real mess here.
 
Sorry it is not a cube in fact, the sides are rectangular. It is mistakenly called a cube. Only the top and bottom are squares.
 
To continue where I left off, I plug H into C= 28xh+8x^2+16x+20h. (h=17000/x^2).

I get this:

sss.png

Considering I am not allowed to use calculus to solve this, I have to graph it or use the vertex formula to find minimum. Something tells me this isn't right, the values are huge and I also can't graph it... (by the way, the final equation can be simplified and it is a fourth degree equation... still doesn't help me)
 
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mickey222 said:
To continue where I left off, I plug H into C= 28xh+8x^2+16x+20h. (h=17000/x^2).

I get this:

View attachment 25142

Considering I am not allowed to use calculus to solve this, I have to graph it or use the vertex formula to find minimum. Something tells me this isn't right, the values are huge and I also can't graph it... (by the way, the final equation can be simplified and it is a fourth degree equation... still doesn't help me)
Click to expand...
Since the equation you get is fourth degree, the "vertex formula" for a quadratic function won't apply. I can't imagine how you'd solve this without either calculus or some technology such as a graphing calculator. Huge numbers are no problem, but high degree is.

Maybe you need to show us the exact wording of the problem as given to you (which is what we ask you for anyway), so we can see if there is some misinterpretation or some information about what you can use. What topic are you currently studying?
 
Dr.Peterson said:
Since the equation you get is fourth degree, the "vertex formula" for a quadratic function won't apply. I can't imagine how you'd solve this without either calculus or some technology such as a graphing calculator. Huge numbers are no problem, but high degree is.

Maybe you need to show us the exact wording of the problem as given to you (which is what we ask you for anyway), so we can see if there is some misinterpretation or some information about what you can use. What topic are you currently studying?
Click to expand...

The design for a rectangular box is being put together which will have a volume of 17,000 cubic inches. Both the top and the bottom are square, and the material for the top and bottom will cost 4 cents per square inch. Each side is made of a different material which will cost 7 cents per square inch. The square top and bottom of the box is fused to the sides with adhesive lining which costs 2 cents per linear inch. The sides are fused with each other with adhesive that costs 5 cents per linear inch.
1. What dimensions of the box will incur the least cost?
2. Determine the least cost.

Right now we're studying zeroes of polynomial functions. It wasn't apparent to me that this problem might concern that topic-- I thought it was actually a carry-over from an earlier lesson on optimization word problems relating to quadratic equations, and that we were just given this problem as review. Perhaps, then, synthetic division goes into this.
 
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If you knew how to find a minimum using the derivative, that would involve finding a zero of a polynomial. But it's not one you could find by simple hand methods (e.g. it isn't a rational root); the best way is still technology.

Desmos, for example, will tell you the (approximate) minimum point:

1613573356974.png

Is there any chance that is allowed in your class?
 
Dr.Peterson said:
If you knew how to find a minimum using the derivative, that would involve finding a zero of a polynomial. But it's not one you could find by simple hand methods (e.g. it isn't a rational root); the best way is still technology.

Desmos, for example, will tell you the (approximate) minimum point:

View attachment 25144

Is there any chance that is allowed in your class?
Click to expand...

I am certainly allowed to use technology to graph the function. I am not allowed to use derivatives. What did you plug into desmos to get that line? I tried desmos before I made this thread, and just got one huge straight line.
 
Ah, I see, you set the equation equal to 0, which I did not originally do. Well, I suppose I will go forward with this minimum. If my cost optimization formula is correct... I'm not sure what else I can do. I figured I had goofed the cost formula, but if you think my work is sound... then this must be the solution.
 
mickey222 said:
Ah, I see, you set the equation equal to 0, which I did not originally do. Well, I suppose I will go forward with this minimum. If my cost optimization formula is correct... I'm not sure what else I can do. I figured I had goofed the cost formula, but if you think my work is sound... then this must be the solution.
Click to expand...
No, I didn't set anything to zero! I entered y = ... .

What I had to do was to adjust the scale for the y-axis.
 
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