2025 Level 2 Calculus Paper

Step 1

Write the surface-area constraint

Because the box has no lid, which surface-area equation is correct?

Step 2

Using \(w=2h\), which expression for the length is correct?

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Step 3

Substitute \(w=2h\) and your expression for \(l\) into \(V=lwh\). Which simplified volume function is correct?

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Step 4

Differentiate \(V(h)\) and solve \(V'(h)=0\). Which height do you get?

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Step 5

Using \(h=0.6\), what is the maximum volume?

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Step 6

Use the second derivative test. What does it show at \(h=0.6\)?

Step 7

Now write the full working clearly.

\text{Surface area}=lw+2lh+2wh=4.32 \] \[ w=2h \] \[ l(2h)+2lh+2(2h)(h)=4.32 \] \[ 4lh+4h^2=4.32 \] \[ lh+h^2=1.08 \] \[ l=\frac{1.08}{h}-h \] \[ V=lwh \] \[ V=\left(\frac{1.08}{h}-h\right)(2h)(h) \] \[ V(h)=2.16h-2h^3 \] \[ V'(h)=2.16-6h^2 \] \[ 2.16-6h^2=0 \] \[ h^2=0.36 \] \[ h=0.6 \] \[ l=\frac{1.08}{0.6}-0.6=1.2,\qquad w=2(0.6)=1.2 \] \[ V_{\max}=1.2\times1.2\times0.6=0.864\text{ m}^3 \] \[ V''(h)=-12h \] \[ V''(0.6)=-7.2<0 \] \[ \text{Therefore the volume is a maximum.}

Great work. The maximum volume is \(0.864\text{ m}^3\), and the negative second derivative proves it is a maximum.