What's The Biggest Box That Can Be Cut From This Cardboard?

We want to maximize the volume of the box, so we need to determine our formula for the volume this box. $V(x) = (x)(30-2x)(30-2x)$ The height is $x$ and the other two dimensions are $30-2x$ since whatever we cut from one side needs to also be cut from the other one, as in the image. Now we find the derivative and find a root, which may be a maxima. $V'(x) = 12 (x^2-20 x+75) = 12(x-5)(x-15)$ $\boxed{x=5}$ This may be proven if we notice that the second derivative if negative at $x=5$ $V''(x) = x-10$ $V''(5) < 0$

Volume of a cuboid = L x B x H

since cardboard will be cut by x cm, Volume becomes: __(30 - 2x)(30 - 2x)(x)

which equals: Volume = 4(x^3) - 120(x^2) + 900x

We differentiate the above equation (USE THE CALCULUS APPLIED MAXIMA MINIMA CONCEPT IN THIS SCENARIO)

achieve: V ' = 12(x^2) - 240(x) + 900

Solving this quadratic, we achieve x = 15 or 5

Thinking logically, if we take 15cm as x, we are doing nothing but cutting the whole box's length and width out lol. So we automatically have the answer as x = 5 cm

$x(20-2x)^2=\frac14[4x(20-2x)(20-2x)]\le\frac14\left(\frac{4x+20-2x+20-2x}{3}\right)^3=2000$ from AM-GM.

$x=5,20$ yields the maximum of $2000$ , but $x=20$ is outside the feasible domain so the answer is $5$ .

volume =(30-2x)(30-2x)x differentiate 30-2x= 4x x=5

There are four options 0, 5, 10, 15.

To cut 4 squares with the side length of 0 cm and 15 cm is nat possible.

If we cut out squares from the board having a side length of 5 cm and make a box with the cardboard, the volume of the board will be (30−2×5)×(30−2×5)×5=2000

And if we cut out squares from the board having a side length of 10 cm and make a box with the cardboard, the volume of the board will be (30−2×10)×(30−2×10)×10=1000

So, if we cut out squares from the board having a side length of 5 cm and make a box with the cardboard, the box will have the most volume and also can hold the most number of chocolates.

So, the answer is 5

it's very simple...for the box to get max chocolate it's volume should b maximum. if u take x = 1) 0 then its volume = 30x30x0 = 0 cm^3 2) 5 then volume = (30-2x5)x(30-2x5)x5 = 20x20x5 = 2000 cm^3 3) 10 then vol = (30-20)x(30-20)x10 = 10x10x10 = 1000cm^3 4) 15 then vol = 0 because no base area so correct ANS = 5 that it hope it helps.........:):):):)

when we cut equal squares of length 5cm then height of box=5cm width=length=20cm and in this way volume of box is maximum than all other cases so it will contain maximum chocolates :)

Four options: 0, 5, 10, 15. It's said that you NEED TO CUT somewhere near the corner to form a box. So, 0 is impossible because 0 means you don't cut anything.

Then, only 5, 10, 15 are left. Since we were asked the minimum value, then choose 5. HAHA.

THIS IS ONLY FOR MULTIPLE CHOICE. DON'T TRY THIS METHOD WHEN IT ISN'T.

(1)The expression for the volume of the box is x(30-2x)(30-2x). (2)Expand to give 4x^3-120x^2+900x. (3)Differentiate to give 12x^2-240x+900. (4)Make (3) equal to zero, to find the turning points, therefore the local maximum and local minimum of (2). (5)Divide through by 12 to get 0=x^2-20x+75. (6)Factorize and solve to give x=5, x=15. (7)Since (2) is a cubic with a positive x^3 term, the turning point with the lesser x value is the local maximum, or use d^2y/dy^2 to do it. (8)x=5.

TO FILL MOST CHOCOLATES THE VOLUME OF THE BOX SHOULD BE MAXI. LET THE VOLUME OF THE BOX IX V. THEN V=(30-2x)(30-2x)x [v=f(x)] then f'(x)=0 at x=5,15 and f''(x)=-ve at x=5 so x=5 Vix maximum.

0 &15 options are cancelled as if x=0 no cardboard will form and if x=15 the whole sheet is cutted into 4 equal squares

now if x=5 then area in which we have to keep chocolates is 400 square units and if x=10 then the area is 100 square units. so, if x=5 we will have max. area.

This problem is an example of an optimization problem which means that we have to get the absolute value that will be the value of the side of the corner in order for the box to hold the maximum no. of chocolates. If x is the length of the side of the squares to be cut out, the volume of the box will be: v(x)=(30-2x)(30-2x)(x) Thus, v'(x)=12x^2-240x+900 =12(x-5)(x-15) Note that x has to be less than half the width of the cardboard because v must be positive. this means that x must be included in (0, 15) and hence the only critical number is 5. Using the Second Derivative Test, v"(x)=24x-240 v"(5)=-120 v"(5)<0 Therefore, v has a relative maximum at x=5 and since it is the only relative extremum in the interval (0, 15), x=5 is also the absolute maximum on (0, 15).

Finally, we can conclude that the maximum length of the side of the square to be cut from the cardboard is 5 cm. so that it can hold the most number of chocolates.

Find V (Volume) in terms of x. Put dV/dx = 0 and solve for x. Get the values of second derivative. Observe that the second derivative is negative at x = -5. Accept x = -5. ignore sign.

I solved the problem by the method of calculus.
Volume of the box, V=(30-2x)^2*x =4x^3-120x^2+900x.
The rate of variation of the volume, V with the variation of x can be expressed as
dV/dx=12x^2-240x+900
For the volume to be maximum or minimum, dV/dx=0
i.e 12x^2-240x+900=0
Solving the above equation we get x=15 or x=5
Viewing the figure itself, we see that V=0 when x=15
Thus, for the maximum value of V, x=5

This one is so simple. From the given options, the least no. will be the solution (since higher the magnitude higher will be its area) 0 is not possible, because we cannot make a box without cutting the edges. So, 5 is the best answer.

You have to take the first derivative of algebraic volume and equate to zero to find the value of x.

That question was boring.

30x30=900 but we need the next greatest area so removing squares of 5cm each we get the area as 20x20=400

We have with ourselves a cardboad which has square dimensions and now we want to make it into box by cutting squares from the side. This box can be seen as a cuboid with a top open. So, we have to find the cuboid with maximum volume.

Differentiate the total volume to get the maxima and minima points - X x (30-2X) x (30-2X) , x=>5,0

The objective is to simply find such a value of x for which the volume of the box is maximum.Hence volume = l b h=x*(30-2x)^2. Now to get max. volume dV/dx = 0 . On solving you get x=5.