Differentiation optimisation

Waliah snowman

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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 9B651AD6-650A-41FE-BBA2-2BA3CA59EED3.png
 
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 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
 
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
 
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
 
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 \(\displaystyle \frac{1}{2}xyz \)
 
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
 
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
 
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
 
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