Maximize the Volume of the Topless Box | (Optimization Problem) |

[@jamestalbott4499]

Thank you!

[@PreMath]

You are very welcome! Thanks for the feedback ❤️

[@RylanceStreet]

The problem of maximising the volume of an open-topped box of surface area 108 square units is the same as maximising the volume of a closed top box of surface area 216 square units, and then cutting that box in half. For the closed top box the maximum volume will be when it is a regular cube, with sides of 6 units. So for the open top box of surface area 108 square units the maximum will occur when it is 6 units by 6 units by 3 units high.

[@geralynpinto5971]

Excellent. Thank you. I have learnt much.

[@AzouzNacir]

The volume of the cube is V=x²y and the area is A=x²+4xy=108, hence y=(108-x²)/4x, hence V=x(108-x²)/4, we have V=(-(x+12)(x-6)²/4)+108, hence V is the maximum for x=6 , y=3 and this maximum value V is equal to 108.

[@imetroangola17]

y = (108 - x²)/4x. V = - (x+12)+... one problem is this negative sign to guarantee your answer.

[@AzouzNacir]

I did not understand what you mean, where is the problem?​@imetroangola17

[@imetroangola17]

@AzouzNacir V= (-(x+12)(x - 6)²/4) + 108? Is that really how you wrote it?

[@AzouzNacir]

​@imetroangola17 Yes, where is the problem?

[@imetroangola17]

@AzouzNacir V = ( - (x+12)(x - 6)²)/4 + 108. My question was your conclusion for x=6, but I realized with more caution that the expression - (x+12)(x - 6)² ≤ 0, for all real x>0. Since we want maximum volume, we should have: (x+12)(x - 6)² = 0, this is only possible if x = 6. Sorry for the misunderstanding. Good solution!

[@jimlocke9320]

Alternative to taking the second derivative: Let V = f(x) = 27x - x³/4. f(x) is a third order equation. When f(x) is plotted, using coordinates (x, V), over the full range (both negative and positive values for x and V), it must have either a local minimum and a local maximum or one "saddle point", where the slope is 0 but is neither a local minimum nor local maximum. Since taking the derivative finds 2 points, x = -6 and x = 6, one must be a local minimum and one a local maximum. We note that f(-6) = -216 and f(6) = 108. Since -216 is less than 108, x =-6 must be the local minimum for V and x = 6 the local maximum. Additional checks to make for both this method and taking the second derivative: There are two other situations to check. 1) For x less than the value of x at the local maximum, 6 in this case, check for values of V exceeding the value at the local maximum. Since values of V for x in the range 0 to 6 are between the local minimum and maximum, they must be less than the local maximum. (V does exceed its value at the local maximum at certain x values less than -6, but negative values of x are not valid solutions to this problem.) 2) There can be no values of V at x values greater than that at the local maximum, x = 6 in this case, that exceed V at x = 6. Since f(x) has negative slope for all values of x greater than 6, this criteria is met. Given a different problem, the local minimum could be to the right of the local maximum and there would be positive x values which would produce values of V exceeding that at the local maximum.

[@anatoliy3323]

Mathematical analysis... I like these kinds of tasks. Thank you so much, Professor 💯👍I wish all the best to you!

[@michaelbenzinger2600]

I have been meaning to ask if you could include some puzzles based on cubes, cylinders, cones, or spheres. You beat me to the punch. Thoroughly enjoyed that. The calculus was a nice touch too!

[@davidseed2939]

incidentally the same idea applies in an open circular cylinder, optimal height is half the width. what about other regular prisms.

[@quigonkenny]

Aₛ = x² + 4xy 108 = x² + 4xy 4xy = 108 - x² y = (108-x²)/4x --- [1] y = 27/x - x/4 From [1], we can see that y is positive when 0 < x < √108 (≈10), and V = 0 when x = 0 and when x = √108 (as y = 0), and positive in between, so we need to find the point in that range where the derivative of V wrt x is 0. V = x²y = x²(27/x-x/4) V = 27x - x³/4 V' = 27(1) - 3x²/4 0 = 27 - 3x²/4 3x²/4 = 27 x² = 27(4/3) = 36 x = √36 [ x = 6 ] y = 27/(6) - (6)/4 = 9/2 + 3/2 [ y = 3 ] Of course, all this is assuming the _external_ surface area is 108, since you could argue that for an open box, surface area includes the inside as well.

[@MegaSuperEnrique]

Even though this is supposed to be for all the Maths, I don't think I have seen a derivative on this channel in the past year! Thank you!

[@PreMath]

You are very welcome! Thanks for the feedback ❤️

[@waheisel]

I think it's the first calculus problem on the channel.

[@waheisel]

Interesting that maximum volume is when the height is half the base. And the number of squares on the surface equals the number of cubes in the volume (108)!

[@uwelinzbauer3973]

​@@waheisel I think, if we maximize the volume of a closed box, it probably will turn out to be a cube?

[@waheisel]

​@@uwelinzbauer3973Great question. It makes sense that turning the open box to a closed one would make all the sides equal when maximumizing volume. I added a top with area 36 and recalculated and got y=x=2√6. You are correct.

[@KeithAllen-pg8ep]

The *final* answer to the given problem is 108 cubic units.

[@Law_Abiding_Citizen_ok]

Should have specified “Exterior Surface area = 108”

[@PreMath]

Thanks for the feedback ❤️

[@cyruschang1904]

The surface area 108 should be 2x^2 + 8xy, not x^2 + 4xy, unless you specify “outside surface only”

[@AzouzNacir]

This is an open box, meaning the top face of the cube is not taken into account.

[@PreMath]

Thanks for the feedback ❤️

[@cyruschang1904]

@@AzouzNacir The box is made up of five boards, each has two surface areas, one facing outward and the other inward.

[@cyruschang1904]

@@PreMath Thank YOU

[@AzouzNacir]

In any question when it is not specified it means the external area i.e. the external part of the box​@@cyruschang1904

[@alexundre8745]

Bom dia Mestre

[@PreMath]

Thanks dear❤️

[@blobfish1112]

I calculated it as a closed box and got x=y=3sqrt2 🤪

[@AM-zh2wl]

6√(3) =square root of 108 same answers? XAndY

[@wackojacko3962]

Clickbait! ... thank God I learnt calculus on this journey to ixtlan. 😊

[@PreMath]

Haha, thanks for the feedback! 😊

[@SKASKumar]

This is calculus problem.

[@PreMath]

Thanks for the feedback ❤️

[@davidseed2939]

interesting that the area and the volume are the same

[@PreMath]

Thanks for the feedback ❤️

[@unknownidentity2846]

Let's do an optimization: . .. ... .... ..... The volume V and the surface area S of the topless box can be calculated as follows: V = x²y S = x² + 4xy Since the surface area is given, we can conclude: S = x² + 4xy S − x² = 4xy ⇒ y = (S − x²)/(4x) V = x²y = x²(S − x²)/(4x) = x(S − x²)/4 = (S/4)x − x³/4 Now we have to find the maximum of V(x): dV/dx = 0 S/4 − 3x²/4 = 0 S/4 = 3x²/4 S = 3x² S/3 = x² ⇒ x = √(S/3) The other solution x=−√(S/3) is negative, so it can be rejected. Now we have to check if this value corresponds to a maximum: d²V/dx² = −3*2x/4 = −3x/2 = −(3/2)√(S/3) < 0 So we finally obtain: x = √(S/3) = √(108/3) = √36 = 6 y = (S − x²)/(4x) = (108 − 36)/(4*6) = 72/24 = 3 V = x²y = 6²*3 = 36*3 = 108 Best regards from Germany

[@SkinnerRobot]

Excellent writeup of the solution. Thanks, UnkownIdentity.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@unknownidentity2846]

@@SkinnerRobot Thanks a lot for your kind feedback and best regards from Germany.

[@marcgriselhubert3915]

I do not see an alternative solution.

[@PreMath]

Thanks for the feedback ❤️

[@משהנאון-י6מ]

לא נכון

[@imetroangola17]

The teacher's calculation is correct!

[@LuisdeBritoCamacho]

MY RESOLUTION PROPOSAL : 01) Volume = V 02) Surface Area = S 03) V (X , Y) = X^2 * Y 04) S (X , Y) = X^2 + 4XY 05) Maximaze V, konwing that S must be equal to 108 ; X^2 + 4XY = 108 06) X^2 + 4XY = 108 ; 4XY = 108 - X^2 ; Y = (108 - X^2) / 4X ; Y = (27 / X) - (X / 4) 07) Now let's substitute Y by ((27 / X) - (X / 4)) in the Equation of Volume. 08) V (X) = X^2 * ((27 / X) - (X / 4)) 09) V (X) = 27X - (X^3 / 4) 10) dV / dX (27X - (X^3 / 4)) ; dV / dX = 27 - (3X^2 / 4) 11) Calculating the Maximum of this Quadratic Function : 12) DV / dX = 0 ; 27 - (3X^2 / 4) = 0 13) X = 6 lin un 14) Y = 3 lin un 16) V(max) = 36 * 3 = 108 cub un Thus, MY BEST ANSWER IS : X = 6 and Y = 3

[@PreMath]

Excellent! Thanks for sharing ❤️