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Lisa, a child with strange requirements for her projects, is making a rectangular cardboard box with square bases. She wants the height of the box to be 3 units greater than the side of the square bases. What should the height be if she wants the surface area of the box to be at least 90 square units while using the least amount of cardboard?
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hi adpqadpq
You could call the size of the base 'x' and form an inequality for x squared using this information. So that would answer the question if we are meant to dis-regard any cut off wastage.
But if we have to start with a sheet of cardboard and design a net for the box with minimum wastage this will be harder to determine. Please clarify.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Yeah disregard any wastage
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This seems to be a straight forward problem resulting in quadratic equation.
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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Could you guys give me some help on this problem?
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Hi;
Where are you stuck? Forming the equation? Solving the equation? Did you draw a diagram?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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I'm unsure of how to approach this. Should i plug in numbers and do trial and error? I don't know.
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I have nothing against trial and error when you are stuck. I see lots of people on forums who would rather admit defeat then plug in a few numbers. But to try that you must set up something to plug numbers into? Can you do that?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Could you guys give me some help on this problem?
I thought I'd done that in post 2.
Call the size of the base x. Then the height will be x+3.
The area of the base is x^2 and for a side wall it will be x(x+3). Thus you can add up and get the total surface area. You will get a quadratic, making an up-side-down U shape. The apex must be 90 or more. So set the quadratic equal to 90 and solve. This will give you the optimum x.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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