<ol>
<li><p>a large cube with edges of length 3 inches is built from 6 small blue unit cubesand 21 small white unit cubes. what is the greatest possible fraction of the surface area of the large cube that could be blue ?</p></li>
<li><p>if y= x^3 -x^2 -7x-3, what is the maximum value of y on the interval [ -3, 3] ?</p></li>
</ol>
<p>0 1 5 14 30
the sequence above starts with 0 and the numbers increase by the squares increase by the squares of consecutive integers. what is the 6th no. in the sequence ?</p>
<p>what is the difference between the maximum value of y= -x^4 -2x^3+5 and the minimum value of y= x^4 +x^2 -4 ?</p>
<p>For the first one, the surface area is 54 and you are going to want those blue squares to cover the corners of the large cube. You have 6 blue squares => 6x3=18. That’s 1/3 of 54.</p>
<p>For the second one, the maximum value of y is when x=-1. y=-1-1+7-3=2</p>
<p>The 6th no in the sequence is 55. The numbers increase by the square of the consecutive integers 1 2 3 4 5.</p>
<p>The answer is 9, in my opinion. The maximum value of the first equation is 5 and the minimum value of the second equation is -4. The difference is 9.</p>
<p>Let me know if those are the correct answers (if you have the answers of course).</p>
<ol>
<li><p>Surface area is 54 square units. The optimal solution is to put each of the blue cubes at a corner (vertex) of the large cube. Each blue cube contributes 3 to the surface area, 6 x3 = 18, and the corresponding ratio is 18/54 = 1/3.</p></li>
<li><p>I’d probably graph the function on a graphing calculator. Most graphing calculators should have the option of finding the max/min of a function on an interval. If you don’t have a calculator, you can find the derivative of y, check critical points, and test these with the boundary points.</p></li>
<li><p>6th term is 30+25 = 55. Simple as that.</p></li>
<li><p>You could also graph these functions or take the derivative. However the minimum of x^4 - x^2 - 4 can easily be found without graphing or calculus. Can you find it?</p></li>
</ol>
<p>MITer94 thanks a lot but can u please help how to find the last one
i have a problem in finding maximum and minium value of polynomial with degree higher than 2</p>
<p>@smartyfox if you don’t know calculus, best is just to graph. But here’s a calculus solution:</p>
<p>Given y = -x^4 - 2x^3 + 5, we have dy/dx = -4x^3 - 6x^2. Setting dy/dx = 0 yields x = 0 or x = -3/2. These two x-values are called critical points. You have to test whether these points are local maxima, minima, or neither. In this case it turns out x = -3/2 maximizes.</p>
<p>Honestly, for most polynomials, best is to graph it. I’m pretty sure most graphing calculators have a function that allows you to find the max. in an interval:</p>
<p>max(f(x), x, x1, x2)</p>
<p>where x1, x2 are the endpoints. If you know calculus, you can use it, but be aware that if you have a polynomial of degree 4 or higher, then its derivative is a polynomial of degree 3, and solving cubics in general is not fun.</p>
all are right except the last one it is 10.2 not 9</p>