Differentiation optimisation

[Waliah snowman]

The end faces of a prism are in the shape of a right angled triangle , as shown. the length of the prism is Lcm. The total surface area of the prism is 500cm squared . Calculate the maximum volume of the prism

 

[Waliah snowman]

I need urgent help with this question please

 

[Romsek]

I would assume that [MATH]L=1[/MATH] to start with. You can scale your solutions by [MATH]\dfrac 1 L[/MATH] at the end.

[MATH]SA = h w + h + w + \sqrt{h^2+w^2}\\ V = \dfrac{h w }{2}\\ \text{solve $\nabla (V - \lambda(SA-500)) = 0$}\\ \text{Select the real and non-negative solutions}\\ \text{You're gonna need some software, (or a lot of time)}\\ \text{scale the solutions found above by $\dfrac 1 L$} [/MATH]
There is a trick suggested by the symmetry of the problem.
Symmetric multivariable expressions usually have extremes when the variables are all the same value.
I think you'll find this is the case here and will lead to a considerable simplification of things.

 

[Waliah snowman]

I would assume that [MATH]L=1[/MATH] to start with. You can scale your solutions by [MATH]\dfrac 1 L[/MATH] at the end.

[MATH]SA = h w + h + w + \sqrt{h^2+w^2}\\ V = \dfrac{h w }{2}\\ \text{solve $\nabla (V - \lambda(SA-500)) = 0$}\\ \text{Select the real and non-negative solutions}\\ \text{You're gonna need some software, (or a lot of time)}\\ \text{scale the solutions found above by $\dfrac 1 L$} [/MATH]
There is a trick suggested by the symmetry of the problem.
Symmetric multivariable expressions usually have extremes when the variables are all the same value.
I think you'll find this is the case here and will lead to a considerable simplification of things.

I really don’t understand how to label the this prism tho with what letters so if you could pleases send me a labelled version and could you tell me what you mean by scaling the solutions by 1/L

 

[Waliah snowman]

I would assume that [MATH]L=1[/MATH] to start with. You can scale your solutions by [MATH]\dfrac 1 L[/MATH] at the end.

[MATH]SA = h w + h + w + \sqrt{h^2+w^2}\\ V = \dfrac{h w }{2}\\ \text{solve $\nabla (V - \lambda(SA-500)) = 0$}\\ \text{Select the real and non-negative solutions}\\ \text{You're gonna need some software, (or a lot of time)}\\ \text{scale the solutions found above by $\dfrac 1 L$} [/MATH]
There is a trick suggested by the symmetry of the problem.
Symmetric multivariable expressions usually have extremes when the variables are all the same value.
I think you'll find this is the case here and will lead to a considerable simplification of things.

Also could you please tell me the answer so I can them try to work towards it hopefully this would be really helpful

 

[Waliah snowman]

I would assume that [MATH]L=1[/MATH] to start with. You can scale your solutions by [MATH]\dfrac 1 L[/MATH] at the end.

[MATH]SA = h w + h + w + \sqrt{h^2+w^2}\\ V = \dfrac{h w }{2}\\ \text{solve $\nabla (V - \lambda(SA-500)) = 0$}\\ \text{Select the real and non-negative solutions}\\ \text{You're gonna need some software, (or a lot of time)}\\ \text{scale the solutions found above by $\dfrac 1 L$} [/MATH]
There is a trick suggested by the symmetry of the problem.
Symmetric multivariable expressions usually have extremes when the variables are all the same value.
I think you'll find this is the case here and will lead to a considerable simplification of things.

Also in class we use a method where we make two expressions one for the volume one for the surface area and then we make one of the unknowns a subject using one of the equations so we have only one unknown then we just use differentiation dy/dx equals to zero then work it out so if you could help me on this method

 

[HallsofIvy]

Your prism consists of three rectangles and two triangles. If the base is an "x by y rectangle" and the back has height z and width y then the base has area xy, the back has area yz. The two right triangles have legs of lengths x and z so hypotenuse of length \(\displaystyle \sqrt{x^2+ z^2}\). The third rectangle, the sloped top of the prism, has area \(\displaystyle y\sqrt{x^2+ y^2}\). The two triangles, the sides, both have area \(\displaystyle \frac{1}{2}zy \).

So the total surface area is \(\displaystyle xy+ yz+ y \sqrt{x^2+ y^2}+ yz= xy+ 2yz+ y \sqrt{x^2+ y^2} = 500 \).

For the volume, think of the triangular side, with area \(\displaystyle \frac{1}{2} zy \) extended across the length x so the volume is \(\displaystyle \frac{1}{2}xyz \)

 

[Waliah snowman]

Your prism consists of three rectangles and two triangles. If the base is an "x by y rectangle" and the back has height z and width y then the base has area xy, the back has area yz. The two right triangles have legs of lengths x and z so hypotenuse of length \(\displaystyle \sqrt{x^2+ z^2}\). The third rectangle, the sloped top of the prism, has area \(\displaystyle y\sqrt{x^2+ y^2}\). The two triangles, the sides, both have area \(\displaystyle \frac{1}{2}zy \).

So the total surface area is \(\displaystyle xy+ yz+ y \sqrt{x^2+ y^2}+ yz= xy+ 2yz+ y \sqrt{x^2+ y^2} = 500 \).

For the volume, think of the triangular side, with area \(\displaystyle \frac{1}{2} zy \) extended across the length x so the volume is \(\displaystyle \frac{1}{2}xyz \)

Thank you soooooo muchhh I will try and work on that now

 

[Waliah snowman]

Your prism consists of three rectangles and two triangles. If the base is an "x by y rectangle" and the back has height z and width y then the base has area xy, the back has area yz. The two right triangles have legs of lengths x and z so hypotenuse of length \(\displaystyle \sqrt{x^2+ z^2}\). The third rectangle, the sloped top of the prism, has area \(\displaystyle y\sqrt{x^2+ y^2}\). The two triangles, the sides, both have area \(\displaystyle \frac{1}{2}zy \).

So the total surface area is \(\displaystyle xy+ yz+ y \sqrt{x^2+ y^2}+ yz= xy+ 2yz+ y \sqrt{x^2+ y^2} = 500 \).

For the volume, think of the triangular side, with area \(\displaystyle \frac{1}{2} zy \) extended across the length x so the volume is \(\displaystyle \frac{1}{2}xyz \)

However are there any clues on how to work with so many unknowns

 

[Waliah snowman]

Your prism consists of three rectangles and two triangles. If the base is an "x by y rectangle" and the back has height z and width y then the base has area xy, the back has area yz. The two right triangles have legs of lengths x and z so hypotenuse of length \(\displaystyle \sqrt{x^2+ z^2}\). The third rectangle, the sloped top of the prism, has area \(\displaystyle y\sqrt{x^2+ y^2}\). The two triangles, the sides, both have area \(\displaystyle \frac{1}{2}zy \).

So the total surface area is \(\displaystyle xy+ yz+ y \sqrt{x^2+ y^2}+ yz= xy+ 2yz+ y \sqrt{x^2+ y^2} = 500 \).

For the volume, think of the triangular side, with area \(\displaystyle \frac{1}{2} zy \) extended across the length x so the volume is \(\displaystyle \frac{1}{2}xyz \)

Could u plsss reply to meeee